• Taylor's Theorem.

Thursday, November 13, 2008

Consequences
Differentiability implies continuity. Algebraic combinations of differentiable functions are differentiable. xⁿ is differentiable (when n is a positive integer).

The Chain Rule
Proved very slickly by stuffing one linear approximation inside another and realizing that the consequence was, indeed, the Chain Rule.

The beginnings of a gallery
Everyone, young and old, learned and naive, should have a gallery of functions to be considered when theorems and definitions are considered. The instructor spent a chunk of time identifying good initial candidates for such a gallery, from "ugly" to "nice". Of course, the beauty of the qualities mentioned depend quite a bit on the observer! Writing in progress!

1. The inverse (?) ruler function
   
2. The characteristic function of the rationals
   
3. The ruler function (Dirichlet?)
   
4. x sin(1/x)
   
5. x²sin(1/x)
   
6. Bumps, etc. (due to Emile Borel)
   
7. The exponential function
   

Monday, November 10, 2008

Increasing functions
I discussed increasing functions and how often they can be discontinuous. This is in the textbook. I wrote and gave a warm advertisement for a book: Functional Analysis by F. Riesz and Nagy (also known and pronounced as "rees-naje". Reprinted by Dover and on sale at Amazon for 14.93. A wonderful book!

Mr. Skalit
Discussed several problems in chapter 4 he had prepared with the help of Mr. Kowalick. In particular, we saw that metric spaces were normal.

Differentiability
The instructor defined and began a discussion of differentiability and announced his preference for a definition written in this form: f:[a,b]→R is differentiable at c∈[a,b] with derivative Q if there is a function E defined in a neighborhood of 0 with lim_v→0E(v)=0 so that f(x)=f(c)+(Q)(x-c)+E(x-c). This definition doesn't have any divisions.

Some simple consequences were mentioned but the Chain Rule will be verified next time.

An exam was announced, to be given in two weeks. Ms. Hood volunteered (??!) to present a problem about differentiation, with the help of Mr. Baldwin.

Thursday, November 6, 2008

Finishing the proof
We used compactness to verify uniform continuity. I continued with the attempted (or rather, interrupted) proof. So X is compact, and we have sequences {p_n} and {q_n} with d(p_n,q_n)<1/n and d(f(p_n),f(q_n))≥ε. Now {p_n} is a sequence in a compact metric space. The sequence itself may not converge, but compactness always guarantees that the existence of a convergent subsequence, {p_nj}. We know that there is an x so that lim_j→∞p_nj=x. Notice that the sequence {q_n} is sort of "dragged along" by the corresponding sequence of p_n's. That is, I claim the subsequence {q_nj} also converges, and its limit is x. Why is that? Well, d(q_nj,x)Δ≤d(q_nj,p_nj)+d(p_nj,x). The first term on the right is <1/n_j. For j large, this will be small and stay small. And the second term behaves similarly, because the p-subsequence converges to x. Now f is continuous at x. Therefore given ε>0, we can find a δ>0 so that if d(x,y)<δ then d(f(x),f(y))<ε. But choose J so that for j≥J, both d(q_nj,p_nj) and d(p_nj,x) are less than δ/2. Then d(q_nj,x)<δ so d(f(q_nj),f(x))<ε and d(f(p_nj),f(x))<ε. Again the triangle inequality applies, so d(f(p_nj),f(q_nj))<2ε for j>J. This is a contradiction to how we created the two sequences initially (that is, it will be a contradiction if you allow me to relabel 2ε as ε!). So we are done.

A consequence of uniform continuity?
I proved a rather remarkable result, and this will be especially wonderful in many applications later. I mentioned that a "random" (?) continuous real-valued function on R can be rather bizarre. One explicit well-known example is xsin(1/x) on [0,1]. For another example with even worse behavior, I investigated the Takagi function graphically in the 9^th meeting of the Byrne Seminar on Experimental Math. A picture drawn there is shown to the right. The Takagi function (named for Teiji Takagi (1875--1960)) happens to be an example of a continuous nowhere differentiable function on [0,1], and it is also a function which is neither increasing nor decreasing on any subinterval of [0,1] of positive length. So continuous functions can be weird and wonderful. In view of that, the result here is almost amazing. If you are willing to tolerate only a little bit of error (as little as you would like) you can approximate any continuous function by a piecewise constant step function with finitely many values. David Tall, an English mathematics educator, gives a detailed discussion of the properties of this function (aimed at teachers) here.

To the right is a possible picture. The graph of a continuous function defined on a closed bounded interval is shown. The light blue band around the graph is ±ε deviation from the graph. The red horizontal lines are the graph of an approximating piecewise constant function with finitely many values. It "steps" because the values it assumes are taken on finitely many intervals of positive length. The result is remarkable to me, because no matter how bad (?) the function f can be, we can replace it by something with finite data (!) if you allow me to commit a very small error. In many applications, the step function is too coarse, and we may want piecewise linear or differentiable or ... whatever. But this result is the initial version. So:

Theorem Suppose f is a continuous real-valued function on [a,b], and ε>0. Then there is a function g:[a,b]→R so that the range of g is finite, g^-1(y) is either empty or a finite set of intervals for all y∈R, and |g(x)-f(x)|<ε for all x∈[a,b].
Proof [a,b] is compact and f is continuous, so f is uniformly continuous on [a,b]. Therefore we can find δ>0 so if |f(x₁-x₂|<δ, then |f(x₁)-f(x₂)|<&epsilon. Now find a positive integer n so |b-a|/n<δ. For all integers j between 0 and n-1, we know that |f(a+j(b-a)/n)-f(x)|<ε for x between a+j(b-a)/n and a+(j+1)(b-a)/n. Define g by this "rule": if x is between a+j(b-a)/n and a+(j+1)(b-a)/n, g(x)=f(a+j(b-a)/n). Then we're done!
Advertisement This is magic, and you should realize it. We have developed a considerable amount of technique, and now many results, including what's in this lecture, are "easy" to verify. Once we have the step function, we can (approximately) integrate f, find its (approximate) mean and standard deviation, etc.

d) implies our first inverse function theorem
There are a collection of results which are referred to as inverse function theorems, including, especially, a result about differentiable mappings between open subsets of R^k. This wonderful result will be proved in 412. One of its standard proofs uses the following result, which is also very frequently invoked in algebraic topology. So: Theorem Suppose f:X→Y is a bijection (1-1 and onto -- basically, X and Y are the same as sets!, f is continuous, and X is compact. Then f^–1 is continuous.
Proof We will use d). To show that f^–1 is continuous, we will investigate if (f^–1)^–1=f has the following property: if C is closed in X, then f(C) is closed in Y. But X is compact, so C, a closed subset, is compact. f(C), a continuous image of a compact subset, is compact in Y. But compact subsets of metric spaces are closed, so f(C) is closed in Y. And we are done!
Advertisement This is magic. There is hardly any effort to verify a result which is used constantly. Be aware, if you know what a topological space is, that the result is true if only X is a compact Hausdorff space.

[0,1)
As I mentioned, NJ State Law QD 324-17 (a math regulation) requires that this example be presented immediately after the statement and proof of the previous theorem.

Consider [0,1) with its usual topology. This is 1 point away (?) from being compact. The mapping f:[0,1)→R² given by f(t)=(cos(2πt),sin(2πt)) is continuous. Considering the picture is more fun. The range of f is x²+y²=1, a closed and bounded subset of R², so the range is compact. But f^–1 is not continuous. Why? The picture shown to the right gives some idea. Take the magenta colored dots in the range, which approach the image of 0 from the wrong way. These dots have a limit, the image of 0. But pulled back to [0,1), we get the green dots which have no limit in the domain of f. Thus f^–1 takes a convergent sequence to a sequence which does not converge, and a continuous function can't do that.

A major result used in calculus
If f:[a,b]→R is continuous, then f([a,b])=[c,d]. This result includes the Intermediate and Extreme Value Theorems. The proof is (now!) easy. A proof from just the definition of continuity is likely to be tedious. So why is this result true? [a,b] is compact and connected, so f([a,b]) is compact and connected. Connected subsets of R are intervals. The only compact intervals are those which are closed and bounded. So we're done.

Connectedness
As mentioned, the lecturer thinks that connectedness as defined is a difficult concept. The definition has a not very prominently mentioned, so it may be irritating to verify that something is connected (or not connected). There is a negative (?) aspect about the logic. So here is a variant.

Arcwise connected
A metric space X is arcwise connected or pathwise connected (both phrases are used) if for all pairs of points p and q in X, there is a continuous function c:[0,1]→X so that c(0)=p and c(1)=q.
Theorem If X is a pathwise connected metric space, then X is connected.
Proof If X is not connected, then X=A∪B with A and B open, non-empty, and disjoint. Take p∈A and q∈B. Then pathwise connected provides a continuous c as in the definition. Since 0∈c^–1(A) and 1∈c^–1(B) and c is continuous, then (using c)!) these are open subsets of [0,1], disjoint, non-empty. But [0,1] is connected, which is a contradiction.

Star-shaped subsets of Rⁿ are arcwise connected, because a star-shaped set has a center, v (or, at least one center!), so that given any p in the set, the line segment from p to v is in the set. So we can connect p to q by detouring (?) through v. Convex sets are star-shaped. Open balls are therefore star-shaped.

For open subsets of Rⁿ...
Much, although not all, of analysis in 411-412 will be in open subsets of Rⁿ. For such subsets, the concepts of arcwise connected and connected coincide.
Theorem Suppose U is an open subset of Rⁿ. Then U is connected if and only if U is arcwise connected.

Proof I did not offer a proof in class, but look: suppose U is connected. I want to show that U is arcwise connected. Fix p∈U. Take A to be the set of all points q which can be "connected" to p with a continuous image of [0,1], as in the definition of arcwise connected. Then A is a connected set, surely. I claim that A is open. Well, if q∈A, then since U is open, there is r>0 with N_r(q)⊂U. I can connect from p to any point in N_r(q) with a detour through q. So A is open. Also, A is closed, since a limit point of q's, call it v, has a ball N_s(v)⊂U. At least one q is in N_s(v), so go from p to that q and then, inside the convex ball, to v. Since A is open and closed and non-empty, A must be all of the connected set U. (Or else what is left out is open, also, and makes U disconnected!)

Not the other way!
For unfriendly sets, arcwise connected and connected may be different. For example, the topologist's sine curve is connected but not arcwise connected. There is a purported proof in the linked page, without much details and also without a picture! Here is a Wikipedia reference, with a picture, but not even a candidate for a proof.

I sketched a rather clumsy proof. My effort would involved repeated use of the Intermediate Value Theorem.

Monday, November 3, 2008

All except the last were previously shown as equivalent. The last is an easy consequence of set and function manipulation results, using fact that a set is closed exactly when its complement is open. The great freedom we now have is to investigate and diagnose results about continuous functions (those which satisfy any/all of the previous properties) and use whichever characterization we like. Some are more convenient to use than others in certain situations.

Composition of continuous functions
We're given f:X→Y and g:Y→Z both continuous. Use c). Take U open in Z, then g^–1(U) is open in Y, and f^–1(g^–1(U)) is open. Now notice that (gof)^–1(U) is the same as f^–1(g^–1(U)).

Distance as a continuous function
We saw that d(x,y₁)–d(x,y₂)≤d(y₁,y₂) because (Δ≤) d(x,y₁)<d(y₁,y₂)+d(x,y₂). Then, switching y₁ and y₂, we know |d(x,y₁)–d(x,y₂)|≤d(y₁,y₂). Therefore if f(y)=d(x,y), f satisfies a Lipschitz estimate, and there must be continuous. See what follows, please.

Lipschitz and locally Lipschitz
A function f:X→Y is Lipschitz if there is a constant, A≥0 so that d(f(x₁,x₂)≤A d(x₁,x₂) for all x₁, x₂ in X. Such a function must be continuous. Use, for example, a). Given ε>0, take δ=ε/A (hey, talk to me if A=0 and you can't figure out what to do).
We'll say f:X→Y is locally Lipschitz if it is Lipschitz in a neighborhood of every point. That is, given x∈X, there is r>0 and A≥0 (A may depend on r and x) so that for all x₁, x₂ in N_r(x), d(f(x₁,x₂)≤A d(x₁,x₂). Such functions are also continuous (same proof!).

Lipschitz conditions are fundamentally important in the standard statement of the existence and uniquenss theorem of ordinary differential equations.

MVT implies Lipschitz
This is jumping ahead a bit, but if f:R→R is differentiable, then the Mean Value Theorem implies that |f(b)–f(a)|≤|f´(c)| |b–a| for at least one c in the interval between a and b. Therefore if we "happen" to know that f´ is bounded on that interval, then f will be Lipschitz there, and we know a Lipschitz constant (the sup of the absolute values of the derivative). Of course, a differentiable function will automatically be continuous, but this maybe gives some feeling for what Lipschitz means: the mapping "stretches" distances by no more than a factor of A, the Lipschitz constant.

Hölder and locally Hölder
Here is a similar idea. Suppose A≥0 and α≥0. Then f:X→Y satisfies a (uniform) Hölder condition if d(f(x₁,x₂)≤A (d(x₁,x₂))^α. For example, the function f:R→R defined by the formula f(x)=sqrt(|x|) satisfies such a condition with α=1/2. It does not satisfy a Lipschitz condition on all of R.
An analogous definition can be made for locally Hölder.

Functions satisfying Hölder conditions are also continuous. For example, take δ=(ε/A)^1/α. The "average" continuous function on [0,1] is not differentiable, but I think it does satisfy, in most places, a Hölder condition of order 1/2. So this is important in many fields (e.g., probability, math finance, physics [Brownian motion], etc.).

Algebraic combinations
Sums, products, and quotients of complex-valued continuous functions are continuous. Probably the simplest way to verify this is by appealing to similar statements about convergent sequences and using the criterion in b) above.

Vector-valued functions and their components
Suppose F=(f₁,f₂,...,f_k):X→R^k. Then each component, f_j, is a function from X to R. And F is continuous if and only if each of the f_j's is continuous. The easiest way to see this is probably to contrast the Euclidean metric, L², in R^k with the L¹ and L^∞ metrics, and use the constants we have previously obtained in problem #3 of Homework #2 to show the needed implications. This is also in the text, of course.

Perturbing the definition of continuous
Suppose f:X→Y. A version of the official definition of continuity follows:

∀p∈X∀ε>0∃δ>0 if x∈X with d(x,p)<δ then d(f(x),f(p))<ε.

∀p∈X∃δ>0∀ε>0 if x∈X with d(x,p)<δ then d(f(x),f(p))<ε.

What about this?
Here's another change to the definition.

∀ε>0∃δ>0 if x,p∈X with d(x,p)<δ then d(f(x),f(p))<ε.

An example
An instructive example is f:R→R defined by f(x)=x². We can show that this function does not satisfy the previous logical statement. We decided that we needed to produce an ε>0 so that for any δ>0 there are points p, q in X with d(p,q)<δ with d(f(p),f(q))<ε. Then we quickly declared that this would be proved if we choose ε=1, say, and created, for each positive integer n, points p_n and q_n so that |p_n–q_n|<1/n but |(p_n)²–(q_n)²|≥1.

The marvelous intervention of Ms. Slusky allowed us to choose p_n=n and q_n=n+(1/[2n]). We computed |(p_n)²–(q_n)²| and saw that it was greater than 1. The instructor then drew a picture intending to show that the graph of this f tilted more and more as |x|→∞ so that getting control over |f(x)–f(y)| even if the size of |x–y| was restricted became more difficult. For more pictures, please see the link below.

A definition
f:X→Y is uniformly continuous if

∀ε>0∃δ>0 if x,p∈X with d(x,p)<δ then d(f(x),f(p))<ε.

Some other examples, perhaps elaborate
Example 1 x² is not uniformly continuous on R.
Comment about learning stuff Especially in math, when the defintions get more and more complicated, a very good idea is to find things which do not satisfy the definition yet are very closely related to things which do satisfy the definition. The differences may be useful in understanding the definition and its consequences.

Example 2 MVT again: conside the function f:R→R defined by f(x)=(3/[4+x²])+17x+cos(4x). I claim that f is uniformly continuous. I'll show this by checking that f satisfies a uniform Lipschitz condition. Then the same δ can be used for any ε and any x. Well, f´(x)=–(6x)/[4+x²]²+17–6sin(x). I claim that each of the "pieces" of this derivative is bounded in all of R. That should be clear fro 17 and –6sin(x), so let's look at (absolute value!) 6x/[4+x²]² for x>0. Uhh ... this is less than 1. Why?
6x≤[4+x²]²: true for x between 0 and 1, certainly, since the largest that 6x gets there is 6, and the smallest that the right-hand side gets is 16. And consider the derivatives:
6≤2([4+x²]2x. The right-hand side is always at least 2(5)2=20 on [1,∞) so it is always larger than 6. Therefore the original functions have the same inequality on [1,∞). This is too much work! It should be easier.

Example 3 5sqrt(x)+17x on [0,∞). Well, the derivative is 5/(2sqrt(x))+17. Mr. Leven correctly objected here because the derivative is not bounded on [0,∞). O.k.: the derivative is bounded on [1,∞) so the function is uniformly continuous there. What about on [0,1]. Aha! Look ahead.

A major theorem
If f:X→Y is continuous and X is a compact metric space, then f is uniformly continuous.

Attempted proof
There's a nice proof in the textbook, using essentially problem 5a) from the first exam. Let me try to give a different proof. What if the theorem were false? Then we would have some ε>0 and two sequences of points {p_n} and {q_n} (remember the verification of the x² example above) so that d(p_n,q_n)<1/n but d(f(p_n),f(q_n))>ε. This alone is not enough to guarantee a contradiction, because we could have such sequences in R and not get into trouble. (In class I gave sequences in R², p_n=(0,n) and q_n=(0,n+{1/n}) but we already had the examples in the x² verification.) What will help us here?

Hint Compactness! This guarantees not that a sequence itself converges, but that some subsequence does. This will be enough to get a contradiction. to be continued ...

Thursday, October 30, 2008

• An absolutely convergent series, ∑_j=1^∞a_j, with ∑_j=1^∞a_j=A and ∑_j=1^∞|a_j|=C.
• An absolutely convergent series, ∑_k=1^∞b_k, with ∑_k=1^∞b_k=B and ∑_k=1^∞|b_k|=D.
• A summation method, that is, a sequence {W_t}_t=1^∞, where
  • Each W_t is a finite subset of NxN.
  • W_t+1⊂W_t.
  • ∪_t=1^∞W_t=NxN.
  and s_t=∑_(j,k)∈Wta_jb_k.

I proved this in detail, perhaps too much detail. The process used the convergence definition applied to the infinite series ∑_j=1^∞a_j and the infinite series ∑_k=1^∞b_k, and the Cauchy criterion applied to ∑_j=1^∞|a_j| and to ∑_k=1^∞|b_k|. The proof used the "technique" of problem 7 in the first exam, and was guided by a version of the rather bizarre picture to the right.

♥Love for absolute convergence♥
The rearrangement result and the product summation result declare that commutativity and associativity are correct when applied in (essentially) any way to absolutely convergent series and to their algebraic manipulations.

Other results about products
I stated Merten's Theorem and another result. Please see the text.

Limits and sequences
I stated the ε–δ definition of limit, and proved that it was equivalent to the sequential statement. This is as in the text. Mr. Baldwin kept me honest.

Continuity
I defined continuity of a function at a point, if the function mapped one metric space to another. Then I defined continuity of a function. I showed that this definition ("Inverse images of open sets are open") was equivalent to requiring an ε–δ statement at each domain/range point pair. This is as in the text. All is an abstract version of the result on real-valued functions defined on the real line which was stated several weeks ago.

Pictures
Ms. Pritsker suggested that I try to show some pictures about what was happening, and I thank her for this.

In this first example, the function shown (from R to R) has what's called a jump discontinuity. The inverse image of the green open interval is traced (backwards) and the result seems to be a half-open interval. So this example fails the "inverse image open" criterion for continuity.

The second example has a more complicated discontinuity. I tried to sketch, in class and here, the function f whose value, f(x), 0 if x≤0 and is sin(1/x) if x>0. Then the inverse image of a small open interval centered around 0 includes (–∞,0] and also includes a countable collection of open intervals, getting smaller and smaller as they "pile up" at 0. Note, though, that although 0 is in the inverse image, there is no open neighborhood (consider the right "half") which is a subset of the inverse image.

The great theorems
Chapter 4 concentrates on precise statements and proofs of some of the most basic theorems in analysis about continuity. These results seem to have their historic source (sort of) in the work of Bolzano. The calculus version of the theorems is the following almost ludicrously simple statement:

Suppose f:R→R is a continuous function, and a≤b. Then the set of values of f([a,b]) is [c,d] with c<d.

Notice that closed bounded intervals in R can be characterized as proper non-empty subsets of R which are both connected and compact. We will prove the result above by showing that continuous images of compact sets are compact, and continuous images of connected sets are connected. Sigh. Both compact and connected are defined in terms of open coverings. So it turns out that the "inverse image open" version of the definition of continuity is exactly suited for quite simple proofs of the results needed. But, please, the apparent simplicity is only superficial. The whole theory is aimed at these results, and 150 years of work have gone into constructing a Definition/Theorem/Proof/Example succession which seems simple and nearly effortless. Historically the results and the ordering involved a great deal of work.

Homework
Please read chapter 4. The top vote-getters in the poll for "Problems students want to do" are Chapter 3: 6, 10 and Chapter 4: 4, 6, 7, 11. I will add one more problem in a formal assignment on Monday.

Monday, October 27, 2008

An example
Here is one (delightful?) example of the sort of computation which made people very uneasy historically. You may remember from the first year of calculus study (geometric series, Taylor series, remainders, etc.) a fact about the Alternating Harmonic Series.

ln(2)=1–(1/2)+(1/3)–(1/4)+(1/5)–(1/6)+(1/7)–(1/8)+...

Now replace each positive term by twice itself minus itself. I hope you can convince yourself this does not change convergence or the sum of the series. Now this is true:

ln(2)=(2–1)–(1/2)+(2/3)–(1/3)–(1/4)+(2/5)–(1/5)–(1/6)+(2/7)–(1/7)–(1/8)+...

ln(2)/2=1–(1/2)–(1/4)+(1/3)–(1/6)–(1/8)+(1/5)–(1/10)–(1/12)+(1/7)–(1/14)–(1/16)+...

Riemann Rearrangement Theorem
If a series of real numbers converges yet diverges absolutely (conditional convergence) then there is a rearrangement which converges to any real number. I gave a proof of this. This is a weaker version of the result in the textbook. Please look there.

The better version in the text
Rearrangements can be devised to have the sequence of partial sums with lim sup and lim inf very arbitrarily specified. There are versions of the theorem which apply to complex series. The conclusions then are more complicated, though.

Summation by parts
We went through the discussion of summation by parts. Several students clearly understood it better than the instructor. Sigh. This was discussed because summation by parts can be applied to several topics in Fourier series.

Some consequences (Alternating series test)
We stated and proved the standard alternating series test.

Products of series
Here we began a discussion of products of series, which is a more complicated undertaking than is immediately apparent. We want to start with two convergent infinite series, ∑_{j=1 (or 0)}^∞a_j and ∑_{k=1 (or 0)}^∞b_k and then analyze the numbers a_jb_k. We'd like to figure out some way to "assemble" these numbers into an infinite series, whose sum, we might hope, is AB, the product of the sums of the two series we began with.

Taylor series and Cauchy products
Thinking about Taylor series almost immediately inspires (?) the definition of the Cauchy product. We "know" from calculus ("know" means that some examples were shown then!) that if f(x)=∑_j=0^∞a_jx^j and g(x)=∑_k=0^∞b_kx^k then a way (?) to organize the product is to suppose it is F(x)=f(x)g(x), a product of functions. And we would hope that F(x)=∑_t=0^∞c_tx^t where now c_t=∑_q=0^ta_qb_t–q. These formulas are gotten in several ways: first, we rewrite the product by assuming that manipulations with infinite polynomials work exactly like those with finite polynomials. Or we can hope that a Taylor's Theorem is at work behind the scenes, and the summation exactly reflects the product rule for l^th derivatives. The sum is sometimes called a convolution of the two sequences. It is unfortunate or, perhaps, interesting, that many of these hopes are not exactly true all of the time we would hope. Oh well. In particular,here is no clear relationship between convergence of the factors and convergence of the Cauchy product. See the example mentioned below.

Example in the text
A neat example in the text shows that the Cauchy product of a convergent series (with itself!) need not converge. Interesting. I have little to add to the account in the textbook.

Dirichlet series and Dirichlet product
Here I defined Dirichlet series, which are rather useful in number theory, and some other areas which don't seem to be immediately related. These are series of the form f(s)=∑_n=1^∞(a_n/n^s). Such series are natural in number theory and complex analysis. The most famous example occurs when all of the a_n's are 1. This is ζ(s)=∑_n=1^∞(1/n^s), the Riemann zeta function. Learn more about this, and make a milliion dollars!. Dirichlet series have a whole theory of their own. For example, just as power series have a radius of convergence inside which they converge absolutely, Dirichlet series similarly (and for much the same reasons!) have an abscissa of convergence, a line x=Constant in the complex plane. The series converges absolutely to the right of that line and diverges to the left of it. For ζ(s), that line is x=1. Or, as we say in the complex analysis game, Re(s)=1.

If we have another Dirichlet series, say g(s)=∑_m=1^∞(b_m/m^s), we might consider F(s)=f(s)g(s) and think about ∑_n=1^∞(a_n/n^s)·∑_m=1^∞(b_m/m^s), and maybe reassemble into another Dirichlet series, so: ∑_N=1^∞(c_N/N^s) and this should be the same as ∑_n=1^∞∑_m=1^∞(a_n/n^s)(b_m/m^s). So we might want c_N/N^s to be, well, which terms of the product? To match up (1/n^s)(1/m^s) with 1/N^s we need nm=N.

Therefore we could define the Dirichlet product of two series (dropping the s stuff) ∑_n=1^∞a_n and ∑_m=1^∞b_m to be the series ∑_N=1^∞c_N where c_N=∑_n·m=Na_nb_m. Here it is divisors which are important. You can see if factoring integers and primality is interesting, such products might be more revealing than Cauchy products. One can then ask if convergence of the factors always implies convergence of the Dirichlet product. The answer, with no further hypotheses, is no. There are examples which don't have the desired behavior.

Absolute convergence and products
We will briefly discussion summation methods for products of series. Let me start with 1 in these series, and starting with 0 would give a similar result. So let's have two series ∑_j=1^∞a_j and ∑_k=1^∞b_k which converge and which have sums A and B respectively. Consider the numbers a_jb_k which I could think of as "sitting" each at a lattice point, NxN, of the plane. Now we will consider a sequence of finite subsets {W_t}_t=1^∞ of NxN, with the following properties:
     1. They are nested: W_t⊂W_t+1.
     2. ∪_t=1^∞W_t=NxN: the union is "everything".

Examples of such W_t's come from both Cauchy and Dirichlet products. Adding up the terms which come from the W_t's exactly will correspond to the partial sums of each of these products. The easy and natural result (I love absolute convergence!) is the following.

Theorem Suppose that the series ∑_j=1^∞a_j and ∑_k=1^∞b_k both converge absolutely and have sums A and B respectively. Then the sequence of (possibly complex!) numbers s_t=∑_(k,l)∈Wta_jb_k converges, and its limit is AB.
Actually, even slightly more is true and follows from the preceding result. That is, the sequence whose t^th term is ∑_(k,l)∈Wt|a_jb_k| also converges, and its value is ≤ the product of ∑_j=1^∞|a_j| and ∑_k=1^∞|b_k|. So any method of summation of the product series of absolutely convergent series "works" and gives the correct answer.

Proof?
We began thinking about the proof and will complete it next time.

A volunteer and his coach
Mr. Skalit, helped by the valiant Mr. Kowalick, will present problems 20, 21, and 22 of chapter 4 in class, probably a week from Thursday.

Problems to be considered
The students in the class will consider Chapter 3: 6, 7, 10; and Chapter 4: 4, 5, 6, 7, 11, 13, 25. Votes for the best problems will be counted, and the 6 or maybe 7 highest vote-getters will be assigned as homework, due on Thursday, November 26.

Thursday, October 22, 2008

The article I mentioned in class which taught me new things about the Ratio Test is this: The mth Ratio Test: New Convergence Tests for Series and is written by Sayel A. Ali. It appears in the June-July 2008 issue of the American Mathematical Monthly.

I started to discuss ♥Why I love absolutely convergent series!♥ My first reason was that rearrangements of such series also converge, and converge to the same sum. This is nice. I didn't quite finish the proof, and I will give an interesting example next time, and a result of Riemann's which explains what happens when the series does not converge absolutely.

Thursday, October 16, 2008

The lecturer continued in a more pedestrian fashion. He defined infinite series. He talked about translating various sequence statements to infinite series: the Cauchy criterion, series with non-negative terms, absolute convergence, comparison, p-series via the Cauchy condensation theorem, and a brief anticipation of a version of the ratio test. More next week!

Monday, October 13, 2008

Inequalities and limits
The only Euclidean space R^k where an order topology coicides with the usual Euclidean metric topology is R¹, that is, R, the real numbers. Some special results follow from using the order in R. For example, we have this theorem:

Theorem Suppose {x_n} and {y_n} are two real convergent sequences with limits L_x and L_y, respectively. If for all positive integers n we know that x_n≤y_n, then L_x≤L_y.
A proof of this was shown, using contradiction. If L_x>L_y, then take L_x–L_y=2ε is positive. Take n large enough so that |x_n–L_x|<ε and |y_n–L_y|<ε. Then –ε<x_n–L_x<ε and –ε<y_n–L_y<ε. So we know that –ε<L_x–x_n<ε (multiply the first inequality by –1) so that x_n–ε<L_x<x_n+ε. Similarly, –y_n–ε<–L_y<–y_n+ε so that (x_n–ε)+(–ε–y_n)=(x_n–y_n)–2ε<L_x–L_y<(x_n–y_n)+2ε=(x_n+ε)+(–y_n+ε). This is already a problem since L_x–L_y=2ε.

The instructor asked if a converse to this result was true, and was greeted with scorn. Indeed, no, since 0<0 but the first 0 could be the limit of the sequence {–1/n} and the second 0 could be the limit of the sequence {1/n}. Huh. But how about this:

Theorem Suppose L_x is the limit of the sequence {x_n} and L_y is the limit of the sequence {y_n}. If we know that L_x<L_y, then there is a positive integer N so that if n>N, x_n<y_n.

The proof of this is that (taking 2ε=L_y–L_x) if |x_n–L_x|<ε and |y_n–L_y|<ε again, then unroll the absolute values as before, you will see that the result quoted is true.

Comment about SUBTRACT INEQUALITIES?
Notice that 1<2 and 1<5 but 1–1<2–5 means 0<–3 which is certainly false! Generally, you cannot subtract inequalities and be sure that the result is correct.

More subtle results occur, but we need a few more definitions.

∞ and –∞
Suppose {x_n} is a real sequence which has the following property: for all M∈R, there is a positive integer N_M so that if n≥ N_M, then x_n>M. Then we will say that lim_n→∞x_n=∞.
If we changed "x_n>M" in what's above to "x_n<M" then we will write lim_n→∞x_n=–∞.

lim sup and lim inf
Suppose {x_n} is a real sequence. Define S to be the set of all subsequential limits of this sequence. This is a collection of numbers in R^*, which I guess is R∪{∞}∪{–∞}, and we will think of R^* as an ordered set with the (more or less) obvious ordering. I asked if we could find examples of sequences with S=∅ and S=R^*. Well ...

NO!
How to think about this: a sort of bisection method. Now consider [–∞,∞] and split it into two parts, [–∞,0]\cup;[0,∞]. Look at N: let's call an integer n left if x_n∈[–∞,0] and call it right. Hey, if it is both left and right call it both. In any case, N is now the union of a set of left and right sets, so one of them has infinitely many elements. Let's suppose it is the right one. Then consider [0,∞]=[0,1]\cup;[1,∞]. Hey: left/right again. If left triumphs (has infinitely many!) then split up [0,1] into [0,1/2]\cup[1/2,1] ETC.. If right wins, then split [1,∞] into [1,2]\cup[2,∞] ETC. Eventually we will get a finite limit or a subsequence which is pushed to infinity. In either case, S≠∞. Whew! So there must be at least one subsequential limit.

YES!
There is a sequence with S=R^*. Take {x_n} to be any "enumeration" of the rationals: so this refers to a function f:N→R which is 1–1 and whose range is Q. Then S is R^*. Why? Because in any real interval of postive length there are infinitely many rationals, so given any such interval and given any integer J, there is always j>J with x_j inside the designated interval. It is not difficult to use this observation to verify that S is indeed R^*.

Question Is there a sequence which has S=R? You think about this, please.

lim sup and lim inf
Given {x_n} consider S, the set of all subsequential limits. Then lim sup is the sup of S (considered as a subset of R^*) and lim inf is the inf of S.

Some special sequences
The sentence "And now we begin." ends the novel, "Portnoy's Complaint", by Philip Roth. This sentiment could be echoed by many analysts at this point in the course. We now start looking at some concrete examples of sequences and then series.

The examples
We saw as a consequence of our study of the Archimedean property, that

• {1/n} converges and its limit is 0.
• If |x|<1, then {xⁿ} converges and its limit is 0.

The lecturer falls, and no student picks him up!
We tried to consider the sequence {a^1/n} for a>0. This sequence does converge, and its limit is 1. After some confusion regarding the Bernoulli inequality, little progress was made towards the verification of the convergence and limit claim. Sigh. Deferred to Thursday's class, I suppose!

Thursday, October 9, 2008

Writing in progress!
He continued with a clumsy proof that the set of subsequential limit points of a sequence in a metric space is always a closed set. There is a tiny bit of diagonalization needed in this proof. The lecturer obfuscated this as much as he illuminated it.

Cauchy sequences
A major discussion was started with the definition of Cauchy sequence, a very important concept in analysis and, indeed, in all of mathematics. Given {x_n} in the metric space (X,d), we call this sequence Cauchy if given any ε>0, there is a positive integer N_ε so that if n and m are integers ≥N_ε, then |x_n–x_m|<ε.

Cauchy seuqences are candidates for convergent sequences. They are defined, though, only with "internal", self-referencing criteria, so that knowing such a sequence must converge (as it must, under certain circumstances!) is neat. In R and Rⁿ Cauchy sequences must converge. The convergence of many algorithms is guaranteed using the Cauchy criterion (that is, by proving that the sequences produced using the algorithm are Cauchy). We looked at some examples in R and some conditions in R.
Example If {x_n} is a real sequence and we know that |x_n–x_n+1|<1/n, then such a sequence need not converge and such a sequence need not be Cauchy. Here the example (reaching for our knowledge of calc 1) is x_n=∑_j=1ⁿ1/j, the sequence of partial sums of the harmonic series. This sequence does not converge (hey: x_2ⁿ≥1+[(n–1)/2] with an easy argument, so the sequence is not even bounded). Why is this sequence not a Cauchy sequence? Well, we need to compare x_n and x_m when n and m are both "free" and large. If we make n large with m>n, the triangle inequality will only provide a bound like this:
|x_n–x_m|≤|x_n–x_n+1|+...+|x_m–1–x_m|<∑_j=n^m1/j, and that sum is not bounded.
Example The metric space Q has Cauchy sequences which do not converge. Any convergent sequence in a metric space is Cauchy (we'll state that formally in a few microseconds). Now take a rational sequence which converges to sqrt(2) in R. It is Cauchy in R, and hence in Q (the metric is the same). But it can't converge in Q since there is nothing for it to converge to. (If it converge to w in Q, the same inequalities would make it converge to w in R. But then since the limits are unique, sqrt(2) would be in Q!)
Example Look at (0,1), the open unit interval in R. The sequence {1/n} is in (0,1). In R it converges to 0, and thus it must be Cauchy, but it can't converge to anything in (0,1) since otherwise it would converge to w>0. But the sequence converges to 0 in R, so (just as before) this is impossible.

A convergent sequence is Cauchy
Use a triangle inequality argument.

If a subsequence of a Cauchy sequence converges, then the original sequence converges.
Use a triangle inequality argument.

A Cauchy sequence is bounded
Indeed, take ε=1. Then the whole "infinite tail" of the sequence after the N_ε term is within distance 1 of x_Nε. Now take a ball big enough to enclose also the finitely many other points left out.

The Direichlet problem, a historical digression
This is a problem in partial differential equations which originated in classical mathematical physics. Attempts to solve this problem in the late 19^th centuries led to recognition that "closed and bounded" is not enough to guarantee convergence in many natural "function spaces", specifically function spaces which were used to analyze the Dirichlet problem. This failure or, rather, perhaps, perceived deficiency led to recognition of the importance of compactness, and to the invention of the notions and methods that we are currently discussing.

We went on ...
The lecturer really deliberately tried to investigate the connection between Cauchy and compactness in a rather diffuse non-linear way. This is often how new mathematics develops!. The sequence of ideas went like this:

• Suppose {x_n} is a candidate Cauchy sequence in a metric space. I would like to understand why it should (or should not!) converge. So I said that convergence is actually not affected by any "initial segment" of the sequence. It is a property of the collection of "infinite tails". So let K_N be the set {x_N,x_N+1,x_N+2,...}. If the sequence converges to some limit L, then L might not be in K_N -- that would be too much to hope for. But certainly if a sequence has a limit, say L, then the limit will be in the closure of any tail of the sequence. Why? Because any ε ball around the limit will have infinitely many terms of the sequence. If these terms are equal to L, then the limit is in K_N. If none of the terms are L, then indeed the limit is inside the closure of the tail.
• Let's call L_N the closure of the tail, K_N. Well what do we know? Certainly, these sets are nested: L_n+1⊂L_n. And how about their sizes? I will measure size with diameter, the sup of the distance between pairs of points of the subsets involved. There are two lemmas:
  • Given ε>0, there is N so that diam(K_n)<ε for n≥N. This is true because of the definition of Cauchy.
  • If S is a subset of a metric space, the diameter of the closure of S is the same as the diameter of S. This is true with a mostly standard sup argument. I hinted at this in class, and it is in the text.
• Now consider the intersection of the L_n's. Suppose that this intersection is not empty. Then the following is true:
  • The intersection has at most 1 point. Because if it has two distinct points, these points have a positive distance between them. We can choose ε>0 smaller than that distance. Then for large enough n (preceding remark!) the L_n will not be able to have both of the points in it.
  • If the intersection is non-empty, the unique point (by the previous observation) in it is the limit of the sequence. This is because given ε>0, we can take n large enough so that the diameter of L_n and K_n is less than ε. But then the points in K_n, an infinite tail of the sequence, are all within ε of the limit, since L is in all of the L_n's.
• Why should the intersection of the L_n's not be empty? In R we could consider L_n=[n,∞), a nested collection of closed sets with empty intersection. Or we could consider (0,1/n), a nested collection of open sets with empty intersection and getting smaller. But these are the same examples we used before when we were considering compactness. In fact, we need to use the fact that in a compact set, the nested intersection property holds. So if all of the L_n's are in a compact set, the intersection of all of them is non-empty. We proved that. Therefore,

  Theorem Any Cauchy sequence in a compact metric space converges. (A direct result of the finite intersection property.)
  Theorem Any Cauchy sequence in R^k converges. (Because any such sequence is bounded, and is therefore inside a very big (maybe!) compact set, and therefore can be consider to be in a compact set, and we are done.)

The textbook has a much more linearly ordered and careful presentation.

Monday, October 6, 2008

We then proved (as in the text) that sums of complex-valued convergent sequences are convergent to the sums of the respected limits. Also, if a sequence converges to a non-zero complex number, then "eventually" the sequence is non-zero, and that the sequence obtained by taking reciprocals of the elements of that tail of the original sequence converges, and the limit is the reciprocal of the limit of the original sequence. These techniques are basic and must be part of all mathematicians' subconscious.

I looked at the product of two metric spaces. We put a metric on this product, defined in a way very similar to the Euclidean metric in R² from d(x,y)=|x–y| in R¹. I observed, very briefly, that metrics giving the same topology on the cartesian product are the analogues of the L¹ and L² and L^∞ metrics mentioned in the last homework assignment. Then we saw that a sequence in the product converged if and only if the components in each factor (X and Y, respectively) converges. The same is true in any finite product, but with infinitely many factors things get much more complicated. (The last part of the second problem of the Entrance Exam deals with similar obstacles.).The metrics which we considered here (L¹ and L² and L^∞) and which give, in finite dimensions, the same topologies and the same notion of sequence convergence, all turn out to be different in infinite dimensions. This is either distressing and bewildering, or wonderfully enriching!
We will use the observations about products of metric spaces almost always with R^k (where k is a positive integer).

We then moved on to subsequences. I tried to be very careful about defining a subsequence, and admitted my occasional confusion between a sequence (a mapping from the ordered set N to X) and the image of the mapping, a subset of X. Here is a weird example.

Example
Since Q, the rationals, are countable, there is a bijection F:N→Q. Fix any such bijection for this discussion. Now fix any x∈R. I claim there is a subsequence G:N→Q of F so that G converges to x. Why? Well, we must write G as FοI where I:N→N is strictly increasing. I will create I inductively as follows.

1. What is I(1)? Consider (x,x+1), an interval of positive length. I(1) is a positive integer n so that F(n) is in (x,x+1). This is true since the rationals are dense.
2. Now we specify I(n) for n>1 inductively. We assume that I(1),I(2),...,I(n–1) have already been defined, with I(1)<I(2)<...<I(n–1). Consider the interval (x,x+(1/n)). There are infinitely many rationals in this interval, else we can easily get a contradiction to the density of the rationals (please: you can and should describe how to get in that case an interval of positive length with no rationals). Since there are infinitely many rationals, there will be a rational q with F(j)=q and j larger than I(n–1). Let I(n) be equal to that j. This inductively defines a subsequence, and clearly, if m≥N, |G(m)–x|<1/n. So the subsequence converges to x.

So the "structure" of subsequences can be quite complicated. I don't think this should be too surprising since there are, after all, uncountably many subsequences.

We stated a result something like this:
Theorem Suppose {x_n} is a sequence in a metric space. The following are equivalent:

1. {x_n} converges to x.
2. All subsequences of {x_n} converge to x.
3. All subsequences of {x_n} converge, and they all must converge to the same x.

Thursday, October 2, 2008

He began chapter 3, defining convergence of sequences in metric spaces. He showed that a convergent sequence was bounded, etc. He proved that the product of convergent sequences of complex numbers converged, and its limit was the product of the limits of the factor sequences.

Monday, September 29, 2008

The Heine-Borel Theorem
If E is a subset of R^k, then the following statements are equivalent.

1. E is closed and bounded.
2. E is compact.
3. Every infinite subset of E has a limit point in E.

A counterexample (not really!)
This is a rather simple but still instructive example. Suppose X is any set, and d is the discrete metric. That is, d(x,y) is 1 when x≠y and is 0 when x=y. Every subset of X is an open set always (and every subset is a closed set!). The compact subsets of X are exactly the finite subsets. The diameter of every subset of X is at most 1. So if X is infinite, there are certainly many closed and bounded subsets which are not compact. So the H-B theorem is not true in all metric spaces.

I note that most "infinite-dimensional" situations which arise in analysis (and, indeed, even in certain areas of algebra) naturally many subsets do not satisfy the H-B Theorem. Certainly compactness implies closed and bounded in a metric space, and several textbook homework problems show that the third condition implies compactness in any metric space. But generally the friendly hypothesis of "closed and bounded" will not imply compactness.

Proof of H-B
This was very much like what's in the textbook, with only minor variations introduced by the instructor to ... uhhhh ... challenge the students. Yeah, that's right: challenge the students.

Perfect sets are uncountable
Thanks to Ms. Pritsker and Ms. Hood. Here is her writeup.

The most wonderful perfect set, the Cantor set
I tried to analyzed this set informally. It is an uncountble set. The intervals taken out of it from [0,1] have total length equal to 1, so maybe it should have total length 0. The points in the Cantor set are those whose ternary (base 3) expansions have no 1's. The Cantor set and variations of it are used as the foundation for many disturbing and unintuitive examples in analysis and topology.

It would be nice to tell you that the Cantor set and its relatives are unnatural, etc., except that certain dynamical systems which closely model physical and biological systems have certain types of behavior closely tied to the Cantor set!

Homework due on Thursday, October 9
The third homework assignment is due a week from Thursday.

Out of town
I will be away from Thursday, October 2, to Sunday, October 5.

Thursday, September 25, 2008

Compact
Define of open cover and compact using a finite subcover.

Compactness turns out to be of fundamental importance in many numerical computations. This is not at all clear from the definition and discussion we're about to do. We are beginning at the tail end of a century of struggle with how to handle compactness and related notions, and the "perfection" shown is a bit difficult for a novice to grasp and certainly to understand in a meaningful way.

Examples
We verified from the definition that [0,1] is compact.
Here's how Suppose {G_α} is an open cover of [0,1]. Let S={s∈[0,1] : [0,s] has a finite subcover by elements of {G_α} }. Then:

1. S≠∅ since 0∈S because 0 is in one of the G_α, and that one open set itself is a finite subcover of [0,1].
2. S is bounded above, since S⊂[0,1] shows that 1 is an upper bound.
3. If w=sup S, then w∈S. This is true because w is in some G_α, and for some ε>0, (w–ε,w+&epsilon)⊂G_α. If w=sup S, there must be an element of S in (w–ε,w+&epsilon) or else w is not a least upper bound. Let v be that element. Then [0,v] has a finite subcover, and so by including G_α in the subcover also, [0,w] has a finite subcover. So w∈S,
4. The sup of S must be 1. Just as in the previous section, we see if w=sup S<1, take G_α, and for some ε>0 with (w–ε,w+&epsilon)⊂G_α. If w<1, then including this G_α with the cover tells us that [0,max(1,w+(1/2)ε)] also has a finite cover, which contradicts that w is an upper bound of S.

We verified from the definition that (0,1) is not compact.
And why this is true As Mr. Baldwin suggested, cover (0,1) by (1/n,1) for each positive integer n. Then the Archimedean property implies this is an open cover (every positive number is greater than some 1/n) and the union of every finite subcover is exactly its "highest" element, and none of the (1/n,1)'s is equal to (0,1).

Balls are enough to check compactness
We need only use open balls instead of open sets in the covers we use to test for compactness.

2.33
Suppose K⊂Y⊂X. Then K is compact relative to X if and only if K is compact relative to Y.
I think we used balls here, which might be slightly nicer than the proof of the text. Suppose X=R² and Y=R¹ (considered as a subset of R² with y∈Y ↔ (y,0)∈X) and with both sets having topologies determined by the usual metric. Notice that the metric in R² defined by sqrt((a₁–a₂)²+(b₁–b₂)²) is just |a₁–a₂| when we consider points on the horizontal axis. K should be a subset of Y.

Now suppose we have a cover of K by open balls in Y and we know that K is compact in X. The picture to the right is an effort to illustrate the situation. I drew a collection of green intervals. I had to lift them up a little bit ("up" means vertical motion) to show them and not have them overlay and conceal K. If we use the same centers and the same radiuses and create open balls in X, then, since the green intervals cover K, the resulting magenta discs will also cover K. This is because the intersection of each magenta disc exactly equals the green interval which it specified. If we know that K is compact in X, then a finite union of the magenta discs will have K as a subset. The related green intervals will have K as a subset, also, since a point in K is in a magenta disc exactly when it is in the related green interval. We have proved that if K is X compact then it is Y compact.
Now suppose we have a cover of K by open balls in X and we know that K is compact in Y. So K is in a union of magenta discs. Take one such disc. Its intersection with Y is not necessarily an open ball in Y. It must be an open subset of Y, because each point in the disc is an interior point (we proved that open balls are open!). Therefore each point in the intersection of a magenta disc with Y is contained in a green interval -- that is, an open disc in Y with center in Y and positive radius. This may be a big collection of green sets, but I don't care: the collection of all of these green intervals, for all of the magenta discs, is an open cover of K in Y. Since K is compact in Y, there's a finite subcover by the green intervals. Each green interval comes from one of the magenta discs. So for each green interval in the finite subcover, take the associated magenta disc. The resulting collection of magenta discs is a finite subcover in X of K. We have proved that if K is X compact then it is Y compact.

2.34
Compact subsets of metric spaces are closed.
Proof as in the text. This is a brief and clever proof and the ideas are used elsewhere.

2.34
Closed subsets of compact sets are compact.
Proof as in the text.

Corollary
If F is closed and K is compact, then F∩K is compact.

2.36
If {K_α} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of {K_α} is nonempty, then ∩K_α is nonempty.
What's written in the hypothesis is called the Finite Intersection Property. One of my colleagues remarked that this result is no more than DeMorgan's Laws. I have always found this confusing. So, with the help of many charitable students, I tried to state the compactness definition carefully and logically. I then wrote the contrapositive. I then used DeMorgan's Laws to translate the statements we obtained. If we consider everything as a subset of one of the given K_α's, we proved the theorem.
This is the proof in the text, where it is given not so histrionically.
Bonus vocabulary word: histrionic Of, or relating to actors or acting; Excessively dramatic or emotional.

Corollary
If {K_n} is a sequence of nonempty compact sets such that K_n⊃K_n+1 (n=1,2,3,...), then ∩₁^∞K_n is not empty.
Proof as in the text. People sometimes say "the K_n's are nested" instead of "K_n⊃K_n+1".

2.37
If E is an infinite subset of a compact set K, then R has a limit point in K.

Proof as in the text.

2.38
If {I_n} is a sequence of intervals in R¹ such that I_n⊃I_n+1 (n=1,2,3,...) then ∩₁^∞I_n is not empty.

Useful and relevant examples

1. Suppose I_n is (0,1/n). Then these intervals are non-empty and nested, and their intersection is ∅. But these intervals are open.
2. Suppose I_n is [n,∞). Then these intervals are non-empty and nested and closed, but unbounded. Their intersection is ∅. But these intervals are unbounded.

slightly restated 2.38
Make the intervals both closed and bounded. Then the result is true. This phrase has great "resonance" historically.
If I_n=[a_n,b_n], then a_n<b_m when n and m are positive integers. If α is sup{a_n :n∈N} and if β is inf{b_m: m∈N} then (not totally easy exercise) α<β, and then [α,β] is ∩₁^∞I_n, and certainly the interval is not empty.
Proof as in the text.

2.39
Let k be a positive integer. If {I_n} os a sequence of k-cells such that I_n⊃I_n+1 (n=1,2,3,...) then ∩₁^∞I_n is not empty.
We briefly discussed what "k-cells" were.

Here are pictures of 1- and 2- and 3-cells. A k-cell is the Cartesian product in R^k of k closed and bounded intervals, one in each factor.
The proof as in the text.

Monday, September 22, 2008

This question is more suitably part of Math 441, but we briefly discussed certain desirable (?) properties possessed by metric spaces but not by "random" topological spaces. More generally in the history of topology, a great deal of effort went into discovering which topological spaces occured as a result of a metrics. This is called metrizability and the theorems are intricate.

Uniqueness of limits/Hausdorff
If a topology comes from a metric, then the topology has the following property, which may initially seem rather strange.
Property H Take two distinct points x and y of the space X. Then there are open sets U and V of X with x∈U and y∈V with U∩V=∅.
To see why this is true in a metric space, let q=d(x,y) which must be a positive number since x and y are different. Then put r=(1/2)q, and consider N_r(x) and N_r(y). If z is in both of those sets, then d(z,x)<r and d(z,y)<r, so (triangle inequality and symmetry of d) q=d(x,y)≤d(z,x)+d(z,y)<(1/2)q+(1/2)q=q which is a contradiction since q is positive.

A silly example of a topology which does not have Property H is the following: take X to be a two-element set, say {♣,♦}. The topology on X (which must a set of subsets of X, don't sink into the sea of abstraction here) to consider is {∅,{♣,♦}}. This sure is a topology (called the indiscrete topology, the smallest topology -- the largest topology is called the discrete topology) and it does not have "enough" open sets to separate the two points, ♣ and ♦.
Comment The number of topologies on a finite set is interesting and there's no known exact pattern. Note, oh combinatorists, that most (!!) of them do not have Property H above.

A topology which has Property H is called Hausdorff (the points can be "housed off" from each other [sorry]). This property is already evident in calculus, because we tell people there the following result: if {x_n} is a sequence and if lim_n&rarrow;∞x_n=A and lim_n&rarrow;∞x_n=B, then A=B. This "uniqueness of limits" property is exactly a consequence of Property H or Hausdorffness. There are important examples of non-Hausdorff topological spaces which seem less "artificial" than what I gave, but their descriptions are somewhat complicated.

Countable sequence of open sets
A topology which comes from a metric has the following property:

Sequence of neighborhoods
If x∈X, then there is a sequence of open sets, {U_n} so that if V is open and x∈V, there must be a positive integer N so that x∈U_N⊂V. Actually, we can even ask that these sets be "nested", so that U_n+1⊂U_n for all n.
Proof A proof is very brief. If d is the metric, take U_n to be the ball of radius 1/n centered at x, that is, U_n=N_1/n(x). If V is as described, there is some r>0 with N_r(a)⊂V. Then the Archimedean property guarantees the existence of N with 0<1/N<r, so that U_N⊂V.

This idea will be used again and again in this course. It plays an important part of the "a implies c" proof and you should look at that now if you already haven't. We will use the idea to "create" suitable sequences. But let me show you a somewhat intricate example of a topological space which does not satisfy the preceding result.

R^∞
As a set, R^∞ is "just" the collection of all real sequences. So if x∈R^∞, then x=(x₁,x₂,...,x_n,...) with all of the x_n's elements of r. For the purposes of this paragraph, I'll call such an x positive exactly when all of the x_n's are positive real numbers. I want to define the box topology on R^∞. I will first begin by defining box neighborhoods.

Suppose x=(x₁,x₂,...,x_n,...) is a point in R^∞ and ε=<ε₁,ε₂,...,ε_n,...) is a positive element of R^∞. Then I will tell you what is in the subset M_ε(x) of R^∞. A point y=(y₁,y₂,...,y_n,...) of R^∞ is in M_ε(x) if and only if, for all positive integers n, |x_n–y_n|<ε_n. The sets M_ε(x) play a role similar to what the balls N_r(x) do in the case of metric spaces, but there's no metric visible (indeed, there is no metric possible, as we will see!). Now I will tell you what an open set. A subset W of R^∞ will be open in the box topology if:
      for all x∈W, there is a positive ε in R^∞ so that M_ε(x)⊂W.
We should check that the rules for a topology are satisfied. If you wish to do this, you should begin by verifying that M_ε(x) is itself open (analogous to verifying that an open ball is open). This is not too difficult, and, indeed, nost of the other verifications are easy. One thing needs to be said, I think. If M_εa(x) and M_εb(x) are two of the defining neighborhoods, and if we put ε_c;=(min(ε_a1,ε_b1),min(ε_a2,ε_b2),...,min(ε_an,ε_bn),...) (take minimums in each coordinate) then M_&epsilon3(x)⊂M_εa(x)∩M_εb(x). This result is needed to prove that intersections of box open sets are box open.

Now suppose that x∈R^∞. What if there were a qualifying sequence of box open sets {U_n} behaving nicely as in the Sequence of neighborhoods proposition above? Then there must be (definition of box open) a sequence of positive ε_n's so that x∈M_&epsilonn(x)⊂U_n. I will create a specific M_η(x) which is not contained in any of the M_&epsilonn(x)'s.

Let me begin by looking at a specific example, because that will help us understand the general case. Take U_n to be M_{(1/n,1/n,...,1/n,...)}(0). So this is a neighborhood of 0 whose "polyradius" is 1/n in each coordinate or direction. Certainly, there is no point other than 0 in all of the U_n's, so they seem to "shrink down" to 0. As a response to this choice of a sequence of U_n's, I ask you to consider V=M_{(1,1/4,...,1/n²,...)}(0). I claim that none of the U_n's can be included inside V. Why? Well, let's look at a point p_n in U_n. Define p_n to be (2/(3n),2/(3n),...,2/(3n),...) so it is 2/(3n) in each coordinate. This is certainly inside U_n since 2/(3n)<1/n. But it is not in V, because 1/n² is eventually less than 2/(3n).

Now let's try something similar in general. You "challenge" me with a list of M_εn(0)'s.
My M_η(0) will be constructed so that its sequence of coordinates →0 faster (eventually!) than any of the sequences of coordinates of the ε_n's. Here is one possible recipe:
     Take η_j=(2/3)min(1/j,ε_1j,ε_2j,...,ε_jj)
Then what do I know? I know that &eta_j is much smaller than the j^th coordinates of ε₁, ε₂, ..., ε_j. The element of R^∞ defined by two-thirds of &epsilon_n is in M_εn(0), surely (I just mean, multiply the components of &epsilon_n each by 2/3). But it is not in V. Why? Because for j>n, η_j is less than (2/3)ε_nj. This "tail" restriction prevents the element of U_n from being in V.

What's going on?
Given a sequence of sequences of positive numbers, we have created a sequence whose limit is 0 and which approaches 0 eventually faster than any of the sequences. The "eventually" is that it is smaller than the given sequences at a variable starting place, of course. This is again the diagonal process applied in a more complicated way.

Please note that in class I used a much simpler prescription to create η which won't necessarily prove what is needed. I am sorry. I also owe emphatic thanks to Ms. Slusky whose persistant inquiries and messages made me think a bit more about this. It is more subtle than what I hurriedly did in class.

First countable
A topology which does satisfy the Sequence of neighborhoods proposition is called first countable. I showed that a topology obtained from a metric space is first countable, and gave an example of a topological space which is not first countable. There are simpler examples but the example given is also Hausdorff, and the underlying set, R^∞ is neither artificial nor unimportant, but arises naturally in probabilistic reasoning (select a random sequence of real numbers). The box topology is not used in probability because it has other defects besides lack of first countability. In a first countable space, sequential reasoning is adequate to determine what is and is not an open set.

Metric spaces are Hausdorff and first countable, and I likely never refer to these terms officially again in this course, but I will use the properties they define frequently.

More definitions
Now for the actual progress of the course. We return to metric spaces,

Interior point; interior of a set
We defined interior point and the interior of a set. We prove that the interior of a set is equal to the union of all the open sets which are a subset of the set. We investigated some examples.

Closed set
A closed set is the complement of an open set. Some sets are neither open nor closed. We investigated some examples. The closure of a set is the intersection of the closed sets containing the set. We characterized this using the idea of limit points, and considered (rapidly!) some examples. A closed set is one which contains its limit points.

Perfect set
A perfect set is a set which is its own limit points. Ms. Pritsker, with the help of Ms. Hood, volunteered to prove, some time, that a perfect set in a metric space is uncountable. Thanks to them in advance.

Dense
I did not define "dense" and will need to begin next time with that definition. Sigh.

Thursday, September 18, 2008

The real numbers are uncountable
Here is what I said. This is, more or less, Cantor's second (!) proof of this assertion (more later). Suppose R were a countable set. Then, since subsets of countable sets are countable, the interval [0,1] would be countable. Now supponse [0,1]~N, the positive integers. Then there would be an "enumeration", that is, a way of listing the elements of [0,1] according to their image under a hypothetical bijection with N. So let us enumerate the elements of [0,1], listing each one according to its decimal expansion address.

First element <---> .a11a12a13a14...
Second element <---> .a21a22a23a24...
Third element <---> .a31a32a33a34...
Fourth element <---> .a41a42a43a44...
....

Criticism of this Cantorial proof
Well, the problem is that the decimal "address" is not unique. What do I mean? By analogy, if we talked about a building at the intersection of Fourth Avenue and Avenue A, it is logically possible that the building would have two distinct addresses. It could be called, say, 47 Avenue A and 222 Fourth Avenue. Certainly that occurs with decimals. For example, I know (?) that .379999999...(9's repeating) is a decimal address for a number which also has the decimal adress .38000000...(0's repeating). Well, hw many numbers have that problem? The problems are those numbers with decimal expansion "ending" in an infinite string of 9's. Well, golly, how many such problematic numbers are there? For any finite string of integers, .c₁c₂c₃...c_L I could end with 9's repeated or subtract 1 fron c_L. (This is not exactly correct, since c_N could be 0 but I'm getting tired.) So this countable set of reals has maybe two addresses. Throw them out and apply the Cantor proof (the "Cantor diagonal process"). The result will be an unenumerated real number. So the proof works. Let's follow Professor Rudin's text, though. This will mean using the diagonal process you have already seen at least two more times, and maybe even more. This is not a bad thing, since the idea is really inspired, first-class, etc.

Really, a set which is not countable
Let's consider the sequences whose values are 0 or 1. That is, such a sequence is a function, f:N (the positive integers)→{0,1}. Or, if you are a traditionalist, it is {a_n}_{n in N} where the a_n's are 0 or 1. What if this set (let us call it, temporarily, S) is countable? We enumerate the sequences in S. That is, S consists of f₁, f₂, f₃, f₄, ... and we "diagonalize" to create a sequence which is not enumerated. So g(n) is 1-f_n(n). I wrote this equation and knew this was sort of being exhibitionistic -- I just wanted to show you I could. Sign. What the heck is g? At the positive integer n, is is 1-f_n(n). So (work it out, I have, darn it!) this "switches" the values. That is, if f_n(n) is 1, then g(n) is 0, and if f_n(n) is 0, g(n) is 1. So g can't be any function already counted. The set S cannot be enumerated, and it is not countable.

Hash [huh?]
Let me now try to create an injective function I:S→R. I will be wrong, but, I will try! First, as preface, I will assert the following wonderful fact:
If n is a positive integer and a is non-negative, then ∑_j=1ⁿr^j=(r–rⁿ⁺¹)/(1–r).
Now you may know this as the partial sum of a geometric series, but I know it in Math 411 as a fact which can be confirmed with mathematical induction if r≠1. Well, if I know this fact, then (with a≥0 and 0≤r<1) any such sum is positive and certainly less than r/(1–r). for example, if r=1/2, the sums will all be less than 1.

Let's try the following to create an f. If A={a_n} is a sequence of 0's and 1's, define I(A)=sup{∑_n=1^Na_n/2ⁿ : N is a positive integer}.

There's a bunch of things to check. First, does the set indicated above have an upper bound? The key observation is that
∑_n=1^Na_n/2ⁿ≤∑_n=1^N1/2ⁿ≤1.
So the set involved in I(A)'s definition is bounded above, and therefore it has a sup. Unfortunately, several students pointed out that the Emperor's clothing is indeed lacking. I is not 1-1. It is, after all, just the sequence {a_n} encoded as the binary expansion of a number. And just as decimals have problems, so do these expansions. For example, the sequence {1,0,0,0,...(all 0's)} gets mapped to a set whose only element is 1/2. But the sequence {0,1,1,1,...(all 1's)}: well, we need the sup of ∑_n=2^N1/2ⁿ={1/2}–{1/2^N). Since we proved that 2^N is unbounded, these sums also have sup equal to 1/2. The mapping I is not injective.

How to fix it
Computer science sometimes calls functions like I by the name, hash function (this is part of the special Math 411 Vocabulary Building Project). The hash function classifies elements of a set. Here the set to be "classified" are the collection of 0-1 sequences, and we will classify them by real numbers in [0,1]. When its values of the hash function are the same for different inputs, the result is called a collision. Can we redefine I to avoid collisions?

One suggestion (from Mr. Kowalik?) was to replace 2 by 3 in the definition of I. This will work. In fact, we will use something similar next week in a slightly different context. I wanted to use a weirder idea, because it leads to a wonderful fact mentioned later. So my "suggestion" was to replace 2ⁿ by 2^n!. This means that we are filing the sequences into bins (?) of [0,1] which have widths which shrink much more quickly. Let's now consider two different sequences, say A={a_n} and B={b_n}. Let's assume that N is the first time that they disagree. So a_n=b_n for n<N and, to be specific, a_N=0 and b_N=1. I claim that I, as modified with n! instead of n for the exponent, will have I(A)<I(B): they won't be equal. If I verify this, then I is injective, and, since S is uncountable, we will have verified that R is uncountable. Wow!

Let's make an I(A) as large as possible and an I(B) as small as possible, and compare them. The initial segments appearing in the sum are the same, and if I show what follows, then we're done:
∑_n=N+1^∞1/2^n!<1/2^N!.
Now consider the sum. Its first term is 1/2^(N+1)!=1/(2^N!)^N+1 (repeated exponents are subtle!). What about the ratio between successive terms? Well, the factorials make what's written above not a geometric series. So let me overestimate what is above by a geometric series with initial term 1/(2^N!)^N+1 and ratio between successive terms equal to 2^{(N+2)!–(N+1)!} (this is just the ratio between the first and second terms -- all the other ratios are much less because the exponents are even larger). The exponent is at least 4. So the series given above is less than a geometric series whose first term is 1/2^(N+1)! and whose ratio is 1/4. The sum of that series is [1/(2^(N+1)!]/(1–{1/4}) which is 1/(3·2^(N+1)!–2). Now let me ask about this inequality:
2^N!<3·(2^(N+1)!–2)
     When N=1, this is 2<3·2⁰ which is true.
     When N=2, this is 2²<3·2⁴ or 4<48 which is true.
     When N=3, this is 2⁶<3·2²² which is very true (heh, heh: 64<125,582,912: silly).
The desired inequality is always true, and can be "left to the student" (factorials grow very fast!). So we are done, and we know the reals are not countable. We will see other proofs of this fact, but now let's use it.

Transcendentals ...
Mr. Ratner discussed problems 2 and 3 in chapter 2. He defined algebraic numbers and showed that they were countable, and therefore concluded that the transcendentals, the "other" real numbers, were uncountable. Here is his writeup.
Note For the algebraists among you, please realize that the algebraic numbers are a field. This is not totally trivial, but it is algebra, so I can't prove it, and I don't need to in Math 411.

Opinion and history
I said as I believe, that this sequence of silly trivial observations is truly remarkable. One consequence is that most real numbers, if the word "most" is used in almost any sense, are transcendentals. So a random real number should be transcendental. It is rather difficult to make sense of the word random in an infinite setting, which is one of the real difficulties of probability.

Cantor's proof of the existence of transcendental numbers as described above is the only one I knew until fairly recently, but apparently it was his second proof. Earlier in his life, he published a proof which can be described as constructive. A discussion of this proof is in the article Georg Cantor and Transcendental Numbers by Robert Gray, in The American Mathematical Monthly, Vol. 101, No. 9 (Nov., 1994), pp. 819-832. If you are at a Rutgers computer, you should be able to see a copy of this paper using this link.

Yeah, there are lots of them; now show me one!
I know because people have told me that the special numbers e and π are transcendental. The original proofs of these facts are difficult and long, and even now, a century later, I do not think there are any really easy proofs. It would be nice, since most real numbers are transcendental, to give at least one explicit example of such a number. Here:
∑_n=1^∞1/2^n!.
The transcendentality of this number is a result of Liouville, whose name is also attached to other results in math and physics. The foundation of the rather brief proof is the Mean Value Theorem of one variable calculus, and the strange statement that algebraic numbers which are not rational can't be approximated very well by rationals in a suitable sense. An exposition is here.

We must go on!
The central concerns of the course will frequently be stated in terms using topology. So we should study a bit of topology. We have a whole course, Math 441, devoted to this. I've taught it. Let me outline what my first lecture in that course was about.

How to begin Math 441
My assumption was that not all students were knowledgeable about advanced math and proofs, but that there would be some recall of calculus. I "remembered" some definitions and introduced a definition new to many students.

Definitions

• A sequence of real numbers {x_n} converges and has limit the real number L (written as lim_n→∞x_n=L) if, given any ε>0, there is N so that if n≥N, then |x_n–L|<ε.
• Suppose f:R→R is a function. f is continuous at a if, given any ε>0, there is δ>0 so that if |x–a|<&delta, then |f(x)–f(a)|<ε.
• A subset U of R is open if, given any w in U, there exists a ρ>0 so that the interval (w–ρ,w+ρ) is inside U.

Theorem Suppose f:R→R is a function. The following logical statements about this function are equivalent.

1. If {x_n} is any convergent sequence with lim_n→∞x_n=L, then the sequence {f(x_n)} also converges, and lim_n→∞f(x_n)=f(L).
2. f is continuous at every point.
3. If U is any open subset of R, then f^–1(U) is an open subset of R.

Proof of b implies c
I will try to verify c, and somewhere I will use b. So suppose U is an open subset of R, and w is in U. Since U is open, I know there is ρ>0 so that (w–ρ,w+ρ)⊂U. This interval translates to the statement: all real y's which satisfy |y–w|<ρ (interval statement and inequality statement are logically equivalent). Suppose w=f(a). Then the inequality becomes |y–f(a)|<ρ. Now use b which declares "there exists" (magic!) δ>0 so that if |x–a|<δ then |f(x)–f(a)|<ε. But think about this: it means in turn that the set f^–1(U) includes the numbers x which satisfy |x–a|<δ. But that means (inequality/interval) (a–δ,a+δ)⊂f^–1(U). Think this through: we have produced an open interval contianing a inside f^–1(U), and we can do this for any a inside f^–1(U). Therefore f^–1(U) is open.

There were no other requests until later that afternoon brave Mr. Leven and brave Mr. Kiria came to my office, risking GBH (in English mysteries, GBH is "grievous bodily harm" but I find to my horror that GBH is also a text-messaging abbreviation for "great big hug"). They worked through what I think is a more difficult proof involving sequences. Let me present their proof here.

Proof of a implies c
I need to convince you that the sequential statement implies the open/inverse open statement. In fact, here we will proceed by proving the contrapositive: if the open/inverse open statement is false, then the sequential statement is false.

Let's suppose I know an open subset U of R so that f^–1(U) is not open. Well, this means there's at least one w in U with f(a)=w and one ρ>0 so that (w–ρ,w+ρ)⊂U, so that there is no τ>0 with (a–τ,a+τ) inside f^–1(U). Let's use this complicated statement to create a sequence. If n is a positive integer, take τ to be 1/n. The interval (a–1/n,a+1/n) is not a subset of f^–1(U), so there must be a number x_n in this interval with x_n not in f^–1(U). That is, |x_n–a|<1/n and f(x_n) is not in U. Since f(x_n) is not in U, it also can't be in the interval (w–ρ,w+ρ) because that interval is a subset of U. Therefore we know that |f(x_n)–f(w)|≥ρ.

Putting this all together, we have created a sequence {x_n} so that |x_n–a|<1/n. Therefore (Archimedean property) this sequence converges to a. But w=f(a), and |f(x_n)–f(a)|≥ρ where ρ is some positive number. The sequence {f(x_n)} can't converge to f(a)! And a must be false! We have verified if c is false, then a is false. Therefore a implies c.

YOU TOO can try to prove one of the implications guaranteed by the theorem above. Tell me about it, or ask me about it!

So what people then did ...
Historically, people decided that statement c was a neat way of studying ideas like continuity. It seemed to change inequalities and complicated implications into "simple set theory" (well, the packaging was good, but the difficulties are just hidden, not gone!). So a whole industry was created to understand this sort of approach.

A metric
I defined metric as in the text, commenting that most difficult rule to verify was usually the triangle inequality. I gave two examples. The most important example for Math 411 was the Euclidean metric (square root of the sum of the squares of the differences in the coordinates). The triangle inequality then is a consequence of the Cauchy-Schwarz inequality. The silliest example of a metric is the so-called discrete metric, where d(x,y) is 1 if x≠y and 0 if x=y. Note that d(x,y) will play the role of |x–y| in arguments similar to what we just went through.

More definitions
An open ball or open neighborhood was defined as in the text: N_r(p) is the collection of x's in X with d(x,p)<r. This plays the part of (p–r,p+r). The open balls in Rⁿ are "clear". In the discrete metric, an open ball of radius<1 is just 1 point. If the radius is ≥1, it is everything. An open set is one which contains an open ball centered at each point in the set. The radius, r, of the ball will usually depend on the point selected! The open sets for the discrete metric are all subsets of the set!

• An open ball is an open set. I drew a picture so I could remember how to prove this.
• Any open set is a union of open balls. So locally, if we want to understand open sets, just think of a ball.
• The intersection of two open sets is an open set. Take the minimum of two positive radiuses to get a suitable positive radius. In fact, this extends to any finite intersection. Generally, an infinite intersection of open sets need not be open.
  Example In R with the Euclidean metric, take U_n=(–1/n,1/n). The intersection of this sequence of open intervals (open sets) is just {0} (Archimedean) and a point is not open using the usual metric.
• The union of open sets is open.
• The empty set, ∅, and the whole space are open (∅ because there are no points in it to "check"!).

So use these properties to define ...
A topology on a set X is a set of subsets of X, let's call it Τ, with the following properties:

1. ∅ and X are in Τ.
2. Finite intersections of elements in Τ are in Τ.
3. Any unions of elements of Τ are in Τ.

I'll give some more examples next time, and go through more of chapter 2.

Monday, September 15, 2008

I gave a good proof of the thing I messed up last time.

C
We discussed the complex numbers, C, as R[i]/(i²+1) and as R² with vector addition and a weird kind of multiplication and as 2 by 2 real matrices of a strange kind (complex conjugation is realized as transpose of matrices in this guise -- I was, as usual, incorrect!).

I verified a few of the elementary properties of complex numbers, as in the text. I then started to prove the Cauchy-Schwarz inequality. The method I used might have looked a bit different, but was actually the same, as what was in the text. I considered the function f(t)=||A+tB||² where A and B are vectors in Cⁿ. We knew that f(t)≥0, so when I "expanded" this function using bilinearity, the discriminant of the resulting quadratic was non-positive. This is almost the Cauchy-Schwarz inequality, and I tried to suggest how it would verify the text's version of Cauchy-Schwarz by multiplying one of vectors by the scalar e^iθ for suitable choice of θ (to make the appropriate dot product real and positive).

Rⁿ
I discussed enough of Rⁿ to be sure that we could recognize it as a metric space in a few days. In particular, I got versions of Cauchy-Schwarz and the triangle inequality.

Sizes of sets (cardinality)
It will be useful to have this language as we discuss certain examples. So I defined (with help!) the following terms:

1. A~B means
   A and B have the same cardinality, same size: there is a bijection between them. Then A~A (the identity as a bijection); if A~B then B~A (the inverse of a bijection is a bijection); if A~B and B~C, then A~C (the composition of bijections is a bijection). We will get into serious paradoxes if we consider ~ as an equivalence relation on all sets, however.
2. A is finite if either A empty or A~{1,2,...,n} for some positive integer n.
   For a fixed non-empty finite set, the integer n is unique (takes some proving!).
3. A is countably infinite if A~N, where N is the set of all positive integers. Examples of countably infinite sets include N and Z (all integers) and Q, the rationals. There are various sources for an explicit bijection of Q with N (an enumeration) such as Proofs from THE BOOK by Martin Aigner, Günter M. Ziegler, and K.H. Hofmann. However, this follows as a consequence of the fact that the countable union of countably infinite sets is countably infinite. This "fact" uses a Cantor diagonal argument.
4. No "biggest" set exists! Given a set A, there is a set B so that no surjection (onto mapping) exists, and therefore there is no bijection, so A and B are not the same size. This is proved using a version of the barber paradox. Here the following notation is customarily used: if A is a set, then 2^A is the set of all subsets of A. Why this notation? Well, 2={0,1} (!) and the notation 2^A indicates the mappings from A to {0,1}. A subset T of A corresponds to the function which is 1 on elements of T and 0 otherwise (the indicator or characteristic function of T).
   Proof: Suppose f:A→2^A is a candidate onto mapping. Define T to be the subset of A defined by {x is in A if x is NOT in f(x)}. If T=f(x) for some x, then either x is in T, so x is in f(x) which means x is in T but then (definition of T) x is not in f(x) which is x, or x is not in T, so x is not in T=f(x), so x satisfies the determining property of T and x is in T. Both assumptions lead to contradictions, so no such f exists.
   Of course this now means there is no"largest" set, since we can keep taking "subset of ... subset of ...". The universe of sets can be difficult to understand.
5. A set is countable if it is either finite or countably infinite.