# On Nash’s embedding theorem. A little eulogy.

There are two kinds of mathematical contributions — work that is important to the history of mathematics and work that is simply a triumph of the human spirit. Paul Halmos.

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Reference: A Beautiful Mind by Sylvia Nasar.

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“The discussion of manifolds was everywhere,” said Joseph Kohn in 1995, gesturing to the air around him. “The precise that Ambrose asked Nash in the common room one day was the following: Is it possible to embed any Riemannian in a Euclidean space?”

It’s a “deep philosophical question” concerning the foundations of geometry that virtually every mathematician — from Riemann and Hilbert to Elie-Joseph Cartan and Hermann Weyl — working in the field of differential geometry for the past (19th century) had asked himself. The question first posed explicitly by Ludwig Schlaffi in the 1870’s, had evolved naturally from a progression of questions that had been posed and partly answered beginning in the mid-nineteenth century. First mathematicians studied ordinary curves, then surfaces, and finally, thanks to Riemann, a sickly German genius and one of the great figures of nineteenth century mathematics, geometric objects in higher dimensions. Riemann discovered examples of manifolds inside Euclidean spaces. But in the early 1950’s interest shifted to manifolds partly because of the large role that distorted space and time relationships had in Einstein’s theory of relativity.

Nash’s own description of the embedding problem in his 1995 Nobel autobiography hints at the reason he wished to make sure that solving would be worth the effort. “This problem, although classical, was not much talked about as an outstanding problem. It was not like, for example, the four colour problem.

Embedding involves portraying a geometric object as — or bit more precisely making it a subset of — some space in some dimension. Take the surface of a balloon. You can’t put it on a blackboard, which is a two-dimensional space. But you can make it a subset of a space of three or more dimensions. Now, take a slightly more complicated object, say a Klein bottle. A Klein bottle looks like a tin whose lid and bottom have been removed and whose top has been stretched around and reconnected through the side to the bottom. If you think about it, it’s obvious that if you try it in a three-dimensional space, the thing intersects itself. That is bad from a mathematical point of view because the neighbourhood in the immediate vicinity of the intersection looks weird and irregular, and attempts to calculate various attributes like distance or rates of change in that part of the object tend to blow up. But put the same Klein bottle in a space of four dimensions and the thing no longer intersects itself. Like a ball embedded in three-space, a Klein bottle in four-space becomes a perfectly well-behaved manifold.

Nash’s theorem stated that any kind of surface that embodied a special notion of smoothness can actually be embedded in Euclidean space. He showed that you could fold the manifold like a silk handkerchief, without distorting it. Nobody would have expected Nash’s theorem to be true. In fact, everyone would have expected it to be false. “It showed incredible originality,” said Mikhail Gromov, the geometer whose book Partial Differential Relations builds on Nash’s work. He went on:

“Many of us have the power to develop existing ideas. We follow paths prepared by others. But most of us could never produce anything comparable to what Nash produced. It’s like lightning striking. Psychologically, the barrier he broke is absolutely fantastic. He has completely changed the perspective on partial differential equations. There has been some tendency in recent decades to move from harmony to chaos. Nash says chaos is just round the corner. “

John Conway, the Princeton mathematician who discovered surreal numbers and invented the game of life, called Nash’s result “one of the most important pieces of mathematical analysis in this century.”

It was also, one must add, a deliberate jab at then fashionable approaches to Riemannian manifolds, just as Nash’s approach to the theory of games as a direct challenge to von Neumann’s. Ambrose, for example, was himself involved in a highly abstract and conceptual description of such manifolds at the time. As Jurgen Moser, a young German mathematician, who came to know Nash well in the mid-1950’s put it, “Nash didn’t like that style of mathematics at all. He was out to show that this, to his mind, exotic approach was completely unnecessary since any such manifold was simply a submanifold of a high dimensional Euclidean space.”

Nash’s more important achievement may have been the powerful technique he invented to obtain his result. In order to prove his theorem, Nash had to confront a seemingly insurmountable obstacle, solving a certain set of partial differential equations that were impossible to solve with existing methods.

That obstacle cropped up in many mathematical and physical problems. It was the difficulty that Levinson, according to Ambrose’s letter, pointed out to Nash, and it is a difficulty that crops up in many, many new problems — in particular, non-linear problems. Typically, in solving an equation, the that is given is some function, and one finds estimates of derivatives of a solution in terms of derivatives of the given function. Nash’s solution was remarkable in that the a priori estimates lost derivatives. Nobody knew how to deal with such equations. Nash invented a novel iterative method — a procedure for making a series of educated guesses — for finding roots of the equations, and combined it with a technique for smoothing to counteract the loss of derivatives.

Donald Newman described Nash as a “very poetic, different kind of thinker.” In this instance, Nash used differential calculus, not geometric pictures, or algebraic manipulations, methods that were classical outgrowths of nineteenth century calculus. The technique is now referred to as the Nash-Moser theorem, although there is no dispute that Nash was the originator. Jurgen Moser was to show how Nash’s technique could be modified and applied to celestial mechanics — the movement of planets — especially for establishing the stability of periodic orbits.

Nash solved the problem in two steps. He discovered that one could embed a Riemannian manifold in a three-dimensional space if one ignored smoothness. One had, so to speak, to crumple it up. It was a remarkable result, a strange and interesting result, but a mathematical curiosity, or so it seemed. Mathematicians were interested in embedding without wrinkles, embedding in which the smoothness of the manifold could be preserved.

In his autobiographical essay, Nash wrote:

So, as it happened, as soon as I heard in conversation at MIT about the question of embeddability being open I began to study it. The first break led to a curious result about the embeddability being realizable in surprisingly low-dimensional ambient spaces provided that one would accept that the embedding would have only limited smoothness. And later, with “heavy analysis,” the problem was solved in terms of embedding with a more proper degree of smoothness.

Nash presented his initial, “curious” result in a seminar in Princeton, most likely in the spring of 1953, at around the same time that Ambrose wrote his scathing letter to Halmos. Artin was in the audience. He made no secret of his doubts.

“Well, that is all well and good, but what about the embedding theorem?” said Artin. “You will never get it.”

“I will get it next week.” Nash shot back.

Jacob Schwartz, a brilliant young mathematician at Yale was also working on it concurrently.

Later, after Nash had produced his solution, Schwartz wrote a book on the subject of implicit function theorems. He recalled in 1996:

I got half the idea independently, but I couldn’t get the other half. It’s easy to see an approximate statement to the effect that not every surface can be easily embedded, but that you can come arbitrarily close. I got that idea and I was able to produce the proof of the easy half in a day. But then I realized that there was a technical problem. I worked on it for a month and couldn’t see any way to make headway. I ran into an absolute stone wall. I didn’t know what to do. Nash worked on the problem for two years with a sort of ferocious, fantastic tenacity until he broke through it. “

Week after week, Nash would turn up in Levinson’s office, much as he had in Spencer’s at Princeton. He would describe to Levinson what he had done and Levinson would show him why it didn’t work. Isadore Singer, a fellow Moore instructor recalled:

“He’d show the solutions to Levinson. The first few times he was dead wrong. But he didn’t give up. As he saw the problem get harder and harder, he applied himself more, and more, and more. He was motivated just to show everybody how good he was, sure, but on the other hand, he didn’t give up even when the problem turned out to be much harder than expected. He put more and more of himself into it.”

There is no way of knowing what enables one man to crack a big problem, while another man, also brilliant, fails. Some geniuses have been sprinters who have solved problems quickly. Nash was a long distance runner. If Nash defined von Neumann in his approach to the theory of game, he now took on the received wisdom of nearly half a century. He went into a classical domain where everybody believed that they understood what was possible and not possible. “It took enormous courage to attack these problems,” said Paul Cohen, a mathematician at Stanford University and a Fields medalist. His tolerance for solitude, great confidence in his own intuition, indifference to criticism — all detectable at a young age, but now prominent and impermeable features of his personality — served him well. He was a hard worker by habit. He worked mostly at night in his MIT office — from ten in the evening until 3.00 AM — and on weekends as well, with, as one observer said, “no references but his own mind” and his “supreme self-confidence.” Schwartz called it “the ability to continue punching the wall until the stone breaks.”

The most eloquent description of Nash’s single-minded attack on the problem comes from Jurgen Moser:

“The difficulty that Levinson had pointed out, to anyone in his right mind, would have stopped them cold and caused them to abandon the problem. But Nash was different. If he had a hunch, conventional criticism did not stop him. He had no background knowledge. It was totally uncanny. Nobody could understand how somebody like that could do it. He was the only person I ever saw with that kind of power, just brute mental power.”

# Some problems and solutions V: Topology by Hocking and Young

Exercise 1-24: Let S be a compact metric space and Y have a countable basis. Let the function space $Y^{X}$ be assigned the compact open topology and prove that $Y^{X}$ has a countable basis.

Proof 1-24: Given a continuous map $f: S \rightarrow T$, we know that the functional value $f(x)$ is a continuous map of x. Can we choose a topology for F so that $f(x)$ is continuous function of f? The answer is “yes” and the development which follows is largely due to Ralph H. Fox (1945) (you can google it). For each compact subset C of S and each open subset U of T, let $F(C, U)$ denote the collection of all mappings f in F such that $f(C)$ is contained in U. The family of all such collections F(C,U) is taken as a subbasis for the compact open topology of F. Thus, each member of the basis of the compact open topology is of the form $\bigcap_{i=1}^{n}F(C_{i}, U_{i})$ where each $C_{i}$ is compact in S and each $U_{i}$ is open in T.

Here $F(C, U)$ denotes the collection of all mappins (continuous) f in F ($f: X \rightarrow Y$) which sends each compact subset of S to an open set (a member of basis)(and here it is given that Y has a countable basis) of Y.

Thus, $Y^{X}$ also has a countable basis.

QED.

Exercise 1-26. If X consists of n points with discrete topology prove that $Y^{X}$ is homeomorphic to $Y^{n} = Y \times Y \times \ldots \times Y$, which has n factors.

Proof 1-26: Let $f: X \rightarrow Yn$ and so $Y^{X} = (f_{1}, f_{2}, \ldots, f_{n})$ where each $f_{i}$ is some f. Let $f_{j}(x_{i})=y_{i}$ be one-to-one and onto. As X has discrete topology, we can assign discrete topology, as a natural consequence, to Y also. Hence, $Y^{X}$ is homeomorphic to $Y^{n}$ also.

QED.

Exercise 1-27: If $d(x,y)$ is a metric on a set M, show that

$\rho(x,y) = \frac{d(x,y)}{1+d(x,y)}$ is a metric on M and that the two metric topologies are equivalent.

Proof 1-27:

Clearly, $\rho(x,y)=0$ if $x=y$ and also it is clear that $\rho(x,y) = \rho(y,x)$ (both these follow from definition of M a metric space). (also $\rho(x,y) \geq 0$

Now, let us try to prove the triangle inequality.

Consider $\rho(x,z) = \frac{d(x,z)}{1+d(x,z)} \leq \frac{d(x,y)+d(y,z)}{1+d(x.z)} \leq \frac{d(x,y)}{1+d(x,z)} + \frac{d(y,z)}{1+d(x,z)}$

We now need the following inequality to hold true:

TPT: $\frac{d(x,y)}{1+d(x,z)} + \frac{d(y,z)}{1+d(x,z)} \leq \frac{d(x,y)}{1+d(x,y)} + \frac{d(y,z)}{1+d(y,z)}$

that is, TPT: the following is true:

$(1+d(x,y)+d(y,z)) (1+d(x,z)) \leq (1+d(x,z))(d(x,y)(1+d(y,z))+d(y,z)(1+d(x,y)))$

That is, TPT, the following is true:

$d(x,y)(1+d(x,y)) + d(y,z) (1+d(x,y)) \leq (1+d(x,z)) \times d(x,y) \times (1 + d(y,z)) + (1+d(x,z)) \times d(y,z) \times (1+d(x,y))$

That is, TPT the following is true:

$d(x,y) + d^{2}(z,y) + d(y,z) + d(x,y)d(y,z) \leq (d(x,y)+d(x,y)d(x,z))(1+d(y,z)) + (d(y,z)+d(x,z)d(y,z))(1+d(x,y))$

which holds true because common terms $d(x,y), d(y,z)$ and $d(y,z)d(x,y)$ cancel out and we know that $d(x,y) \leq d(x,z) + d(y,z)$.

Hence the triangle inequality holds true for $\rho(x,y)$ and so $\rho(x,y)$ is a valid distance metric.

Now, we need to prove the equivalence between the two metric topologies:

Consider Theorem 1-4: A necessary and sufficient condition that two bases $\mathcal{B}$ and $\mathcal{B^{'}}$ for two topologies in a set S be equivalent is that if p is a point of an element B of $\mathcal{B}$, then there is an element $B^{'}$ of $\mathcal{B^{'}}$ containing p and contained in B, and conversely.

Let us consider open ball $d(x, x_{0}) and let $x_{0}=p$ and consider the open ball of radius r $B_{r}(x,x_{0})$

To find appropriate $B_{R}^{'}(x,x_{0})$

Consider $\rho(x,x_{0}) = \frac{d(x,x_{0})}{1+d(x,x_{0})} < \frac{r}{1+d(x,x_{0})}= R^{'}$

Hence, $B_{R^{'}}(x,x_{0}) \subset B_{r}(x,x_{0})$

Conversely, let us consider open ball: $\rho(x,x_{0}) < r$, that is, $B_{x,x_{0}}^{'}$ so that we get

$\frac{d(x,x_{0})}{1+d(x,x_{0})} < r$ from which we get that $B_{R}(x,x_{0}) \subset B_{(x,x_{0})}^{'}$ where $R = \frac{d(x,x_{0})}{2+d(x,x_{0})}$.

Hence, these two metric topologies are equivalent.

Exercise 1-28: Let M and N be two metric spaces with metrics d and $\rho$ respectively. Show that the product set $M \times N$ is metrized by

$\delta((x_{1}, y_{1}),(x_{2}, y_{2})) = (d^{2}(x_{1},x_{2}) + \rho^{2}(y_{1}, y_{2}))^{1/2}$

Proof 1-28:

By definition of metric and the square root function, we know that $\delta((x_{1}, y_{1}),(x_{2}, y_{2}))\geq 0$

Also, again by definition of $\rho$ itself and definition of metric, we have $\delta((x_{2},y_{2}),(x_{1}, y_{1})) = (d^{2}(x_{2},x_{1})+ \rho^{2}(y_{2}, y_{1}))^{1/2} = (d^{2}(x_{1}, x_{2})+ \rho^{2}(y_{1}, y_{2}))^{1/2} = \delta((x_{1}, y_{1}),(x_{2}, y_{2}))$ so that the function $\delta$ is symmetric.

Now, let us try to prove the triangle inequality:

that is, want TPT3: $\delta ((x_{1}, y_{1}),(x_{3}, y_{3})) \leq \delta((x_{1},y_{1}),(x_{2}, y_{2})) + \delta((x_{2}, y_{2}),(x_{3}, y_{3}))$,

that is, want TPT: $(d^{2}(x_{1}, x_{3})+ \rho^{2}(y_{1},y_{3}))^{1/2} \leq (d^{2}(x_{1}, x_{2})+\rho^{2}(y_{1}, y_{2}))^{1/2} +(d^{2}(x_{2}, x_{3})+\rho^{2}(y_{2}, y_{3}))^{1/2}$

that it, we want TPT: on squaring both sides:

$d^{2}(x_{1}, x_{3})+\rho^{2}(y_{1}, y_{3}) \leq d^{2}(x_{1}, x_{2}) + \rho^{2}(y_{1}, y_{2}) + d^{2}(x_{2}, x_{3}) + \rho^{2}(y_{2}, y_{3}) + 2(d^{2}(x_{1}, x_{2})+\rho^{2}(y_{1}, y_{2}))^{1/2} \times (d^{2}(x_{2},x_{3} )+\rho^{2}(y_{2}, y_{3}))^{1/2}$

But we also know that because d itself is a metric function, the following inequalities are true:

$d(x_{1}, x_{3}) \leq d(x_{1}, x_{2}) + d(x_{2}, x_{3})$ and $\rho(y_{1}, y_{3}) \leq \rho(y_{1}, y_{2}) + \rho(y_{2}, y_{3})$

On squaring the above two triangle inequalities, the following are also true:

$d^{2}(x_{1}, x_{3}) \leq d^{2}(x_{1}, x_{2}) + d^{2}(x_{2}, x_{3}) + 2d(x_{1}, x_{2})d(x_{2}, x_{3})$ and

$\rho^{2}(y_{1}, y_{3}) \leq \rho^{2}(y_{1}, y_{2}) + \rho^{2}(y_{2}, y_{3}) + 2\rho(y_{1},y_{2}) \rho(y_{2}, y_{3})$ and upon adding these two the following is true:

$d^{2}(x_{1}, x_{3}) + \rho^{2}(y_{1}, y_{3}) \leq d^{2}(x_{1}, x_{2}) + \rho^{2}(y_{1}, y_{2}) + d^{2}(x_{2},x_{3}) + \rho^{2}(y_{2}, y_{3}) + 2d(x_{1}, x_{2})d(x_{2}, x_{3}) + 2\rho(y_{1}, y_{2}) \times \rho(y_{2}, y_{3})$, so if we amalgamate the above two main inequalities, we are now required to show that:

$d^{2}(x_{1}, x_{3}) + \rho^{2}(y_{1},y_{3}) \leq d^{2}(x_{1}, x_{2}) + \rho^{2}(y_{1},y_{2}) + d^{2}(x_{2}, x_{3}) + \rho^{2}(y_{2}, y_{3}) + 2 \times d(x_{1}, x_{2}) \times d(x_{2}, x_{3}) + 2 \times \rho(y_{1}, y_{2}) \times \rho(y_{2}, y_{3}) \leq d^{2}(x_{1}, x_{2}) + \rho^{2}(y_{1}, y_{2}) +d^{2}(x_{2}, x_{3}) + \rho^{2}(y_{2}, y_{3}) + 2(d^{2}(x_{1}, x_{2})+\rho^{2}(y_{1}, y_{2}))^{1/2} \times (d^{2}(x_{2}, x_{3})+\rho^{2}(y_{2}, y_{3}))^{1/2}$.

Now, the above will be true if the following is true (and of course, we are now trying to check it):

$(d(x_{1}, x_{2}) \times d(x_{2}, x_{3}) + \rho(y_{1}, y_{2}) \times \rho(y_{2}, y_{3}))^{2} \leq (d^{2}(x_{1}, x_{2})+\rho^{2}(y_{1}, y_{2})) \times (d^{2}(x_{2}, x_{3}) + \rho^{2}(y_{2}, y_{3}))$

The above is clearly true upon full expansion, as we can see it.

Hence, the triangle inequality holds for the given product metric. So the product metric given here is indeed a distance metric function.

QED.

Cheers,

Nalin Pithwa

# Some problems and solutions IV: Topology by Hocking and Young

Reference: Topology by Hocking and Young, Dover Pub., Inc, Available Amazon India

Exercise 1-21: Let $E_{\omega}$ consist of all sequences $x = \{ x_{n}\}$ of real numbers with the metric

$d(x,y) = \Sigma_{n=1}^{\infty} \frac{1}{n!} \times \frac{|x_{n}-y_{n}|}{1+|x_{n}-y_{n}|}$. First prove that this is a metric function. Then show that a necessary and sufficient condition for the sequence $p_{i}=\{ p_{n_{i}}\}$ to converge to $p = \{ p_{n}\}$ is that $p_{n_{i}}$ converge to $p_{n}$ for each n.

Proof:

Part I: To prove that d is a metric function.

Proof Part I : Considering the first axiom of metric distance function, consider two same sequences in $E_{\omega}$ $x = \{ x_{n} \} = y = \{ y_{n} \}$ $\forall n \in \mathcal{N}$ .

Then, clearly we have in this case $d(x,y)=0$ as $|x_{n}-y_{n}|=0$ when $x=y$

Also axiom 2 of definition of metric (concerning symmetry) $d(x,y) = d(y,x)$ because $|x_{n}-y_{n}| = |y_{n}-x_{n}|$

Now we got to check the third axiom of triangle inequality:

TPT: $d(x,z) \leq d(x,y) + d(y,z)$ where $x, y, z \in E_{\omega}$

PS: Direct proof is v difficult. I used “some reverse engineering approach”. Below is my attempt:

TPT: $\Sigma_{n=1}^{\infty} \frac{|x_{n}-z_{n}|}{1+|x_{n}-z_{n}|} \times \frac{1}{n!} \leq \Sigma_{n=1}^{\infty}\frac{1}{n!} \times \frac{|x_{n}-y_{n}|}{1+|x_{n}-y_{n}|} + \Sigma_{n=1}^{\infty}\frac{1}{n!} \times \frac{|y_{n}-z_{n}|}{1+|y_{n}-z_{n}|}$

That is, TPT: $\frac{|x_{n}-z_{n}|}{1+|x_{n}-z_{n}|} \leq \frac{|x_{n}-y_{n}|}{1+|x_{n}-y_{n}|} + \frac{|y_{n}-z_{n}|}{1+|y_{n}-z_{n}|}$

That is, TPT: $\frac{|x_{n}-z_{n}|}{1+|x_{n}-z_{n}|} \leq \frac{(|x_{n}-y_{n}|)(1+|y_{n}-z_{n}|) + (|y_{n}-z_{n}|)(1+|x_{n}-y_{n}|)}{(1+|x_{n}-y_{n}|)(1+|y_{n}-z_{n}|)}$

That is, TPT: $(|x_{n}-z_{n}|)(1+|x_{n}-y_{n}|)(1+|y_{n}-z_{n}|) \leq (1+|x_{n}-z_{n}|)(|x_{n}-y_{n}| + |x_{n}-y_{n}||y_{n}-z_{n}| +|y_{n}-z_{n}| + |y_{n}-z_{n}||x_{n}-y_{n}|)$

That is, TPT: $(|x_{n}-z_{n}|)(1+|x_{n}-y_{n}|+|y_{n}-z_{n}|+|x_{n}-y_{n}||y_{n}-z_{n}|) \leq (1+|x_{n}-z_{n}|)(|x_{n}-y_{n}|+|x_{n}-y_{n}||y_{n} -z_{n} |+|y_{n}-z_{n}| + |y_{n}-z_{n}||x_{n}-y_{n}| )$

In the above, the RHS =

$|x_{n}-y_{n}| + |x_{n}-y_{n}||y_{n} -z_{n}| + |y_{n}-z_{n}|+ |y_{n}-z_{n}||x_{n}-y_{n}|+|x_{n}-z_{n}||x_{n}-y_{n}|+|x_{n}-z_{n}||x_{n}-y_{n} ||y_{n} -z_{n}| + |x_{n}-z_{n}||y_{n}-z_{n}| + | x_{n}-z_{n}||y_{n}-z_{n} || x_{n}-y_{n} |$

Expanding RHS and cancelling same terms from either side, we are left with:

TPT: $|x_{n} - z_{n} | \leq |x_{n}-y_{n} | + |x_{n}-y_{n}| |y_{n} - z_{n}| + |y_{n}-z_{n}| + |y_{n}-z_{n}||x_{n}-y_{n}| + |x_{n}-z_{n} ||x_{n}-y_{n} ||y_{n}-z_{n} |$

But we know that the following is true property of real numbers: $|x_{n}-z_{n}| \leq |x_{n}-y_{n}| + |y_{n}-z_{n}|$ so obviously the whole above inequality holds true and thus the third axiom holds true.

Part 2: Convergence of the sequence:

Prove that the necessary and sufficient condition for the sequence $p_{i} = \{ p_{n_{i}}\}$ to converge to $p = \{ p_{n}\}$ is that $p_{n_{i}}$ converge to $p_{n}$ for each n.

Let us assume that $p_{i} \rightarrow p$ which means by definition of convergence of a sequence that for any $\epsilon>0$ we have $d(p_{i},p) < \epsilon$ $\forall i \geq N_{0} \in \mathcal{N}$ but note that each $p_{i}$ is a subsequence $p_{n_{i}}$ so the condition boils down to $d(p_{n_{i}}, p_{n}) < \epsilon$ for all $i \geq N_{0} \in \mathcal{N}$ and also when $n_{i} \geq M \in \mathcal{N}$ but n remains constant. Reversing the arguments, we get the converse.

QED.

Exercise 1-22. Let M and N be metric spaces and let $f: M \rightarrow N$ be a transformation. Show that f is continuous if and only if the convergence of a sequence $\{ x_{n}\}$ to a point x in M implies the convergence of ${f(x_{n})}$ to f(x) in N.

Proof 1-22:

Use theorem 1-8 of the text: Let $f: M \rightarrow N$ be a transformation of the metric space M with metric d into metric space N with metric $\rho$. A necessary and sufficient condition that f be continuous is that $\epsilon>0$ for $x_{n} , x \in M$, then there exists $\delta >0$ such that if $d(x,y)<\delta$ then $\rho(f(x), f(y))<\epsilon$

Applying this theorem to given problem, if for any $\epsilon>0$ there exists $\delta>0$ such that if $d(x_{n},x) < \delta$ for some $n \geq N_{0} \in \mathcal{N}$ then $\rho(f(x_{n}, f(x)))<\epsilon$ holds true because the map f is continuous. And by definition of convergence of a sequence. The proof works both ways.

QED.

Exercise 1-23: Suppose that M is a separable metric space and $\{ a_{n}\}$ is a countable dense subset of M. If a is any point of M, let $f(a) = (x_{1}, x_{2}, \ldots, x_{n}, \ldots)$ where

$x_{1} = d(a_{2},a) -d(a_{2},a_{1})$

$x_{2} = d(a_{3},a) -d(a_{3}, a_{1})$

$\vdots$

$x_{n} = d(a_{n+1},a) - d(a_{n+1},a_{1})$

$\vdots$

Prove that f is a homeomorphism of M into the space $E_{\omega}$ of the previous exercise.

Proof Exercise 1-23:

We have to prove that $f: M \rightarrow E_{\omega}$ is a homeomorphism from metric space M to metric space $E_{\omega}$

Part I : TPT: $f: M \rightarrow N$ is continuous.

Proof Part 1:

Once again let us try to utilize the Theorem 1-8: Let $f: M \rightarrow N$ be a transformation of the metric space M with metric d into the metric space N with metric $\rho$. A necessary and sufficient condition that f be continuous is that if any $\epsilon>0$ and x is a point of M, then there is a number $\delta>0$ such that if $d(x,y)<\delta$ then $\rho(f(x), f(y))<\epsilon$.

So let us take any $\epsilon>0$ and let $a, b \in M$ such that

$f(a) = (x_{1}, x_{2}, x_{3}, \ldots, x_{n}, \ldots)$

$f(b) = (y_{1}, y_{2}, y_{3}, \ldots, y_{n}, \ldots)$

Now recall that M is a separable metric and ${a_{n}}$ is a countable dense subset of M. So any point of M is a point of this set $\{ a_{n}\}$ or a limit point of this set/sequence ${a_{n}}$.

Now, we need to find a $\delta>0$ such that if $d(a,b)< \delta$ then we have $\rho(f(a), f(b))<\epsilon$.

That is, given $\epsilon>0$ want $\delta>0$ such that:

IF : $d(a,b) < \delta$….THEN,

$\rho(f(a), f(b)) < \epsilon$ but note that f maps a and b to sequences x, and y in $E_{\omega}$.

$\rho(x,y) = \Sigma_{n=1}^{\infty} \frac{1}{n!} \times \frac{|x_{n}-y_{n}|}{1+|x_{n}-y_{n}|}$. But the sequences $\{x_{n} \}$ and $\{ y_{n} \}$ are as defined In metric space M.

Hence, we now have $\rho(x,y) = \Sigma_{n=1}^{\infty} \frac{1}{n!} \times \frac{|d(a_{n+1},a)-d(a_{n+1},a_{1})+d(b_{n+1},b)-d(b_{n+1}, b_{1})}{1-|D|}$ where we have put $D=x_{n}-y_{n}$

But because M has a countable dense subset $\{ a_{n}\}$ it follows that a and b are limit points of those respective sequences in M. This implies $d(a_{n+1},a)<\delta$ and $d(b_{n+1},b)< \delta$. Therefore the above formulation becomes now:

$\rho(x,y) \leq \frac{1}{n!} \times \frac{|2\delta - d(a_{n+1}, a_{1}) -d (b_{n+1}, b_{1})|}{1-D} \leq \frac{1}{n!} \times \frac{2\delta}{1-D} \leq 2\delta$.

So in the limiting case we have to chose $\delta = \epsilon/2$ and we have prove that f is continuous. QED of Part 1.

Part 2: TPT: $f^{-1}: E_{\omega} \rightarrow M$ is continuous.

Before this we got to check if $f: M \rightarrow E_{\omega}$ is one-one and onto. We assume so. I think that proof of part I can be reproduced with arguments reversed. QED.

QED.

Exercise 1-24: Let X be a compact metric space and let Y have a countable basis. Let the function space $Y^{X}$ be assigned the compact open topology, and prove that $Y^{X}$ also has a countable basis.

Proof 1-24:

Let $f: X \rightarrow Y$ where X is a compact metric space and Y is topological space with countable basis. Let $S = Y^{X}$ be the set of all such functions forming a compact open topology.

# Some problems and solutions III: Topology by Hocking and Young

Reference: Topology by Hocking and Young, Dover Publications Inc., Available Amazon India.

Exercise 1-18: Prove that if A is infinite, then $\mathcal{P}_{A}S_{\alpha}$ has limit points although none of its factor spaces has limit points. (Here: For each $\alpha$ in some index set A, let $S_{\alpha}$ denote a space consisting of two points$x_{\alpha}$ and $y_{\alpha}$ with the discrete topology).

Proof 1-18:

We start with a collection G of spaces and a set $A= \{ \alpha\}$ an indexing set. We select some definite indexing function $\phi: A \rightarrow G$ and designate $\phi(\alpha)$ by $S_{\alpha}$. We wish now to topologize $\mathcal{P}_{A}S_{\alpha}$. Eech of the factor space is finite with discrete topology and has no limit points, so each factor space is vacuously closed. Using Tychonoff topology, we take a set $\mathcal{P_{A}}U_{\alpha}$ to be a basis element if each $U_{\alpha}$ is open in $S_{\alpha}$ and, for all, but a finite number of values of $\alpha$, $U_{\alpha}=S_{\alpha}$. So for all except finitely many values of $\alpha$, each $U_{\alpha}$ is closed in $S_{\alpha}$, and we know that any set is closed if it is equal to its closure, hence if A is infinite, the set $\mathcal{P}_{A}$ has limit points (even though the factor spaces don’t have). Hence, the proof.

QED.

Exercise 1-19. Show that $\mathcal{P}_{A}S_{\alpha}$ has no connected subsets other than its one-point subsets. (Here again, for each $\alpha$ in some index set A, let $S_{\alpha}$ denote a space consisting of two points $x_{\alpha}$ and $y_{\alpha}$ with the discrete topology).

(As of now, I am not clear about the proof of this).

Exercise 1-20. The (middle third) Cantor set is composed of all points in the closed interval $[0,1]$ whose triadic expansion (base 3) contains no units. Show that if A denotes the positive integers, then $\mathcal{P}_{A}S_{\alpha}$ is homeomorphic to the Cantor set.

Proof 1-20: A Cantor set is an interesting example of a closed set on the real line. It can be constructed as follows: Delete the open interval $(\frac{1}{3}, \frac{2}{3})$ from the closed interval $F_{0}=[0,1]$, and let $F_{1}$ denote the remaining closed set, consisting of two closed intervals. Then, delete the open intervals $(\frac{1}{9}, \frac{2}{9})$ and $(\frac{7}{9}, \frac{8}{9})$ from $F_{1}$, and let $F_{2}$ denote the remaining closed set, consisting of four closed intervals. Then delete the “middle third” from each of these four intervals, getting a new closed set $F_{3}$, and so on. Continuing this process indefinitely, we get a sequence of closed sets $F_{n}$ such that

$F_{0} \supset F_{1} \supset F_{2} \supset \ldots F_{n} \ldots$

(such a sequence is said to be decreasing). The intersection

$F = \bigcap_{n=0}^{\infty}F_{n}$

of all these sets is called the Cantor set. Clearly, F is closed (since the intersection of an arbitrary number of closed sets is closed), and is obtained from the unit interval $[0,1]$ by deleting a countable number of open intervals. In fact, at the nth stage of the construction, we delete $2^{n-1}$ intervals each of length $1/3^{n}$.

To describe the structure of the set F, we first note that F contains the points

$0,1, \frac{1}{3}, \frac{2}{3}, \frac{1}{9}, \frac{7}{9}, \frac{8}{9}, \ldots$….call this as ***

that is, the end points of the deleted intervals (together with the points 0 and 1).

However, these are not the only points F contains. F contains many other points. In fact, given any $x \in [0,1]$ suppose that we write x in ternary notation, representing x as a series:

$x = \frac{a_{1}}{3} + \frac{a_{2}}{3^{2}} + \frac{a_{3}}{3^{3}} + \ldots + \frac{a_{n}}{3^{n}} + \ldots$…call this ****

where each of the coefficients $a_{i}$ is zero, one or two only. Then it is easy to see that x belongs to F if and only if x has a representation (****) such that none of the numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ equals 1.

(Note because: just as in the case of ordinary decimals, certain numbers can be written in two distinct ways. For example, $\frac{1}{3} = \frac{1}{3}+ \frac{0}{3^{2}}+\ldots + \frac{0}{3^{n}}+ \ldots = \frac{0}{3} + \frac{2}{3^{2}} + \frac{2}{3^{3}}+\ldots + \frac{2}{3^{n}+ \ldots}$ since none of the numerators in the second representation equals 1 the point 1/3 belongs to F which is already obvious from the constrution of F)

Remarkably enough, the set F has the power of the continuum, that is, there are as many points in F as in the whole interval $[0,1]$ despite the fact that the sum of the lengths of the deleted intervals equals

$\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + \ldots =1$

To see this, we associate a new point $y = \frac{b_{1}}{2} + \frac{b_{2}}{2^{2}} + \ldots + \frac{b_{n}}{2^{n}} + \ldots$

and each point (****) where

$b_{n}=0$ if $a_{n}=0$ and $b_{n}=1$ if $a_{n}=2$ (here remember that if x has two representations of the form **** then one and only one of them has no numerators $a_{i}$ equal to 1. These are the numbers used to define $b_{n}$.)

In this way, we set up a one-to-one correspondence between F and the whole interval $[0,1]$. It follows that F has the power of continuum. Let $A_{1}$ be the set of points ***. Then, $F= A_{1} \bigcup A_{2}$ where the set $A_{2}=F-A_{1}$ is uncountable, since $A_{1}$ is countable and F itself is not. The points of $A_{1}$ are often called “points of F of the first kind” and those of $A_{2}$ are called points of the second kind. Since F can be considered equivalent to its proper subset $A_{1}$, there exists a homeomorphism as follows: If J is the set of positive integers, there exists a homeomorphism from the set $\mathcal{P}_{J}S_{\alpha}$ to the Cantor set. Note that both have the same “dimension”, viz, power of the continuum.

QED

Cheers,

Nalin Pithwa

# Some problems and solutions II: Topology by Hocking and Young

Reference: Topology by Hocking and Young, Dover Publications Inc., Available Amazon India.

Chapter 1: Topological Functions and Spaces:

Exercise 1-13: Show that countable compactness is equivalent to the following condition:

If $\{ C_{\alpha}\}$ is a countable collection of closed sets in S, satisfying the finite intersection hypothesis, then $\bigcap_{i=1}^{\infty}C_{i}$ is non empty.

Proof:

Closely related to compactness is the notion of countable compactness. A topological space S is countably compact if every infinite subset of S has a limit point in S. (The reason for using the word “countable” here is that if every countably infinite set has a limit point, then every infinite set does, because every infinite set contains a countably infinite set). In general topological spaces, this property is not EQUIVALENT to compactness although we do have this equivalence in metric spaces.

Theorem 1-23. A compact space is countably compact.

Now, for the present proof, part I : Let us assume that it is given that a topological space S is countably compact. Then, every infinite subset X of S has a limit point in S. A point p is a limit point of a subset X of S provided that for every open set containing p also contains a point of X distinct from p. As X is infinite, we can choose/create a countable collection $(0)_{i=1}^{n}$ of open sets containing p and each such set contains also contains another point of X distinct from p. Now the complement of each such open set is closed. So we now have a countable collection of closed sets $C_{i=1}^{n}$. Then, clearly a finite intersection of these are non-empty, for they contain at least one point p. But a countable collection $C_{i=1}^{n}$ is a subset of an arbitrary collection of such closed subsets. Hence, the arbitrary intersection $\bigcap_{i}^{\infty}C_{i}$ also has a non empty intersection, viz, at least one point p. Hence, part I is proved that countable compactness implies the given condition.

Part II. We have to prove that the given condition implies countable compactness. Proof: The steps are almost “reverse arguments” of the above proof of part I.

QED.

Exercise 1-14: Prove that a compact subset of a metric space is closed.

Proof 1-14: From Theorem 1-22, a compact topological space is countably compact.

Let us consider the given metric space as a metrizable space S. Hence, a compact subset X of S is countably compact. Hence, every infinite subset X of S has a limit point in X. A point p is a limit point of a subset X of S provided that every open containing p also contains a point of X distinct from p. Hence, a point p is a limit point of a subset X of S if every complement of “the open set containing p and a point distinct from p” does not contain p. Hence, subset X contains all its limit points also. Hence, $X=\overline{X}$. Hence, S is closed. Hence, a compact subset of a metric space is closed.

Exercise 1-16. Is the open interval $(a,b)$ compact ?

Answer 1-16: The given open interval is a subspace of the metrizable space $E^{1}$. From previous problem: if X is a compact subset of a metrizable space, then it is closed (in that metrizable space). That is, if a subset X is not closed in a metrizable, then it is not compact in that metrizable and hence, the open interval $(a,b)$ is not compact.

QED.

# Some problems and solutions: Topology by Hocking and Young

Reference: Topology by Hocking and Young, Dover publications, Inc. Available in Amazon India.

Chapter 1: Topological Spaces and Functions.

Exercise 1-8: Show that the mapping f used in Theorem 1-18 is continuous.

Theorem 1-18 is reproduced below:

Consider the mapping f carrying each point $(x_{1}, x_{2}, \ldots, x_{n+1})$ of $E^{n+1}-0$ onto the point $(\frac{x_{1}}{|x|^{2}}, \frac{x_{2}}{|x|^{2}}, \ldots, \frac{x_{n+1}}{|x|^{2}})$.

We want to prove that the above is continuous. We can use the classical $\epsilon \delta$ definition of continuity.

Let $\epsilon >0$. In the image, this becomes $d(x,y)<\epsilon$ where $x=(\frac{x_{1}}{|x|^{2}}, \frac{x_{2}}{|x|^{2}}, \ldots, \frac{x_{n+1}}{|x|^{2}}) \in E^{n+1}-0$ and $x, y \in E^{n+1}-0$ where $y=(\frac{y_{1}}{|y|^{2}}, \frac{y_{2}}{|y|^{2}}, \ldots, \frac{y_{n+1}}{|y|^{2}})$.

Now, let there be a $\delta >0$ such that $d(x_{i}, y_{i})<\delta$, that is, $\sqrt{\Sigma(x_{i}-y_{i})^{2}}<\delta$, which boils down to $\Sigma(x_{i}-y_{i})<\delta$ or that $x_{i}

Case let $|x|=|y|$. That is the $E^{n+1}$ norms of x and y are equal.

Then, $d(x,y)>\epsilon$ becomes

$\frac{1}{|x||y|} \times (\Sigma(x_{i})|y|^{2} - \Sigma(y_{i})|x|^{2}) < \epsilon$, that is, since $|x|=|y|$, we have

$(\Sigma(x_{i})-\Sigma(y_{i})) \frac{|x|^{2}}{|x|^{2}}>\epsilon$, that is,

$\Sigma(x_{i}) -\Sigma(y_{i})<\epsilon$ when we choose $\epsilon = \delta/(n+1)$.

Case: $|x|>|y|$ (the other case $|y|>|x|$ is just a switch of variables).

let $|x|=|y|+k$ where $k>0$

Then, $d(x,y) < \epsilon$ becomes

$\frac{\Sigma(x_{i})|y|^{2}-\Sigma(y_{i})(|y|+k)^{2}}{|x||y|} < \epsilon$

That is, $\frac{(\Sigma{x_{i}}-\Sigma{y_{i}})|y|^{2}-(\Sigma(y_{i})(2|y|k+k^{2})}{|x||y|}< \epsilon$

In the limiting case, we can consider $\Sigma{x_{i}} = \Sigma{y_{i}} + \frac{\delta}{n+1}$. Substituting in the above, we need $\delta$ such that

$\frac{|y|^{2}\delta/(n+1)-(\Sigma(y_{i}))(2|y|k+k^{2})}{|x||y|}<\epsilon$. That is,

$(\frac{|y|}{|x|} \times \delta \times \frac{1}{n+1} - (\Sigma(y_{i})\frac{2k}{|x|}) -\Sigma(y_{i})k^{2}\times \frac{1}{|x||y|}) < \epsilon$

That is, $\frac{|y|}{|x|} \times \frac{\delta}{(n+1)} < \epsilon +$ some positive number, therefore

$\frac{|y|}{|x|} \times \frac{\delta}{(n+1)} < \epsilon$. That is, we can choose

$\delta = \epsilon \times (n+1) \times \frac{|y|}{|x|}$.

Hence, the proof.

QED.

Exercise 1-9. Given any closed interval $[a,b]$ in $E^{1}$, find a continuous mapping of $E^{1}$ onto $[a,b]$ thereby proving that $[a,b]$ is connected.

Solution :

I think I can offer a somewhat partial solution as follows:

The function $y=f(x) = \frac{2}{\pi}(\arctan {x})$ establishes a homeomorphism between the whole real line and the open interval $(-1,1)$.

If some of the reader can help make the proof water-tight, I would be obliged.

QED.

Exercise 1-10. Show that $E^{n+1}-S^{n}$ is the union of two disjoint open connected sets.

Solution to 1-10. This solution has been given to me by my friend, Dr. Amod Agashe, Florida State University (FSU), Tallahassee.

• The idea is that there is an interior and exterior of $S^{n}$ in $E^{n+1}$. So the sphere $S^{n}$ is the set whose distance from the origin is 1. The interior is the points with distance less than 1 and exterior is the points with distance greater than 1. When n=2, this is just the usual sphere in 3D. I think that if we can use cartesian coordinates to show that in this case, the interior and exterior are open and connected (intuitively obvious, but details can be tedius), then the proof should generalize to higher dimensions.
• QED.

Exercise 1-11. Let P be a hyperplane in $E^{n}$ given by $A_{1}x_{1}+ \ldots + A_{n}x_{n}=B$. Show that $E^{n}-P$ is the union of two disjoint open sets.

Proof to 1-11:

Consider the simplest case n=1. That is, $A_{1}x_{1}=B$. Here now P is just a specific point which is deleted from $E^{1}$ clearly decomposing $E^{1}$ into a union of two disjoint non empty open interval $(-\infty, x_{1})$ and $(x_{1}, \infty)$.

Now consider $n=2$. Then $A_{1}x_{1}+A_{2}x_{2}=B$ where the coefficient parameters are arbitrary, but fixed represents a straight line in the plane $E^{2}$. This line clearly decomposes $E^{2}$ into two open planes $A_{1}x_{1}+A_{2}x_{2}>B$ and $A_{1}x_{1}+A_{2}x_{2}.

Now consider case $n=3$. In this $E^{3}$, $A_{1}x_{1}+A_{2}x_{2}+A_{3}x_{3}=B$ represents a specific plane dividing $E^{3}$ into two disjoint, non empty open sets $A_{1}x_{1}+A_{2}x_{2}+A_{3}x_{3}>B$ and $A_{1}x_{1}+A_{2}x_{2}+A_{3}x_{3}/

And, so on, in $n>3$.

Hence, the proof.

QED.

It is instructive to compare the statement and proof of this question to statement and proof of this Lemma 1-17 which is reproduced here: For n>1, the complement of the origin in $E^{n}$ is connected.

Proof of Lemma 1-17:

The hyperplane P whose equation is $x_{n}=1$ in $E^{n}$ is homeomorphic to $E^{n-1}$. (note also that in general, the dimension of the hyperplane is one less that of the Euclidean space). Let $x=(x_{1}, x_{2}, \ldots, x_{n})$. If x is not the origin, at least one coordinate $x_{j}$ is not zero. If $j \neq n$, the line consisting of all points $(x_{1}, x_{2}, \ldots, x_{j}, \ldots, x_{n-1}, t)$ where t is a real number, contains the point x, intersects the plane P, and does not pass through the origin. (This line is normal to P). If $x_{n}$ is the only nonzero coordinate of the point x, the line of points joining x and the point $(1,0,0, \ldots, 0,1)$, that is, points $[t,0, \ldots, t+(1-t)x_{n}]$ does not pass through the origin, does intersect P, and does contain x. The union of P and all these lines is connected because (Suppose that C is connected subset of a space S and that $\{ C_{\alpha}\}$) is a collection of connected subsets of S, each of which intersects C. Then, the space $S^{'}=C \bigcup (\bigcup_{\alpha}C_{\alpha})$ is connected), and obviously fills up $E^{n}/0$. QED.

Exercise 1-12: Show that the torus (the surface of a doughnut) is the continuous image of $E^{2}$ and is therefore connected.

Proof Ex 1-12:

The Euclidean plane $E^{2}$ is the product of two straight lines. (Note the Euclidean plane $E^{2}$ is open). The torus is the product of two circles. Consider now an interval $[a,b]$ and its product with itself. There exists a one-to-one onto mapping from the “diagonal” of $[a,b] \times [a,b]$ to the circle whose diameter is the diagonal of $[a,b] \times [a,b]$. Let S be the set of all non-negative real numbers with their metric topology and let T be the unit circle in its metric topology. For each x in S, let $f(x)=(1, 2\pi (x)^{2}/(1+x^{2}))$, a point in the polar coordinates on T. Then, f is continuous and one-to-one. But the continuous image of a connected space is connected. Hence the surface of a doughnut is continuous image of $E^{2}$ and is therefore connected.

# Advice to young people from Anthony Bonato, mathematician

1. Go to college.
2. Study math.
3. There is more to math than calculus.
4. There are many infinities.
5. Arithmetic is deep.
6. Graph theory is universal.
7. Go to office hours.
8. Find a quiet place to think.
9. Read everything.
10. Be kind.

# A brief word on history of Topology

Maurice Frechet first considered abstract spaces. The concept of a topological space developed during the following years , accompanied by a good deal of experimentation with definitions and fundamental processes. Much of the development of the theory may be found in Felix Hausdorff’s classic work: Grundzuge der Mengenlehre.

Cheers,

Nalin Pithwa

# Caution: topology vs topological space

Note that not every set with a topology is a topological space. If S is a topological set, then for S to be a topological space, it must be possible to obtain the given topology by selecting certain subsets of S as open sets satisfying the three axioms of a topology and to recover the given limit point relations using these open sets.

Cheers,

Nalin Pithwa

# Another brief word on the significance of Topology

Topology is the area of mathematics which studies topological spaces, that is, spaces where continuity can be defined. But topological spaces can be very weird, and at some point, one may wonder whether spaces with strange topologies serve any purpose, apart from being helpful to understand the theory. Natural spaces are those attached to geometrical problems (that is, about figures that can be drawn and manipulated our space) or physical problems (that is, particles and systems that moving according to some physical laws). In these spaces, we can naturally fix coordinates to locate the figures or the particles (at least in a local region of the space) as happens for the surface of the Earth. This leads to the notion of the manifold as the most important object to study in topology.

Cheers,

Nalin Pithwa.