## Analysis HW, compactness and uniform continuity.

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polymer
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### Analysis HW, compactness and uniform continuity.

I've had a realization recently that math is harder then I thought it was. But I'll save you that drama!

Here's the question, it's problem ten of Rudin in continuity.

Prove that a continuous function on a compact metric space X to a metric space Y is uniformly continuous.
Begin with: Assume there exists an [imath]\epsilon > 0[/imath] and some sequences [imath]\{p_n\}, \{q_n\} \subset X[/imath] such that [imath]d_X(p_n,q_n) \rightarrow 0[/imath] and [imath]d_Y(f(p_n), f(q_n)) > \epsilon[/imath].

And use the fact that all infinite subsets of some compact set have a limit in the compact set.

Progress so far:

I've fleshed out set theoretical definitions of the theorems and assumptions made so far, and I've negated the definition of uniform continuity to see why it implies the above beginning of the proof. That hasn't seemed to help illustrate some obvious symmetry of the problem. I know that the hypothesis is necessary for this theorem to be true, hence that seems to suggest that the range of the chosen sub-sequences should be compact in X. The only condition I have on the sequences is [imath]d_n(p_n,q_n) \rightarrow 0[/imath]. Also the only way I know for a subset of a compact metric space to be compact, is for the subset itself to be closed. So given this initial motivation and understanding, I'm trying to consider what useful relationships I have with closed sets and sequences. Assuming that it is compact though, that would imply that the range of the functions of the sequences should also be compact. But if there is a difference of epsilon for an infinite number of elements, than there can't be a finite subcover of an open cover that only includes either of the elements for infinitely many elements. This presentation is vague, and I'm still not sure how to use the previously mentioned theorem. I've burned four hours on this problem so far and it doesn't feel like it should be that complicated, and I have others I have to move on to. So I've decided to raise a white flag and request for help...Am I reasoning in the right direction?

Qaanol
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### Re: Analysis HW, compactness and uniform continuity.

You’re pretty close, but you don’t need anything about closedness. Sequential compactness will do just fine. I recommend working through it again fresh, on a blank piece of paper. Write out the definitions and negations again, and see if you can come up with something useful. Maybe sleep on it and see if you think of something in the morning.

If you’re really stuck

Spoiler:
Wikipedia has a proof.
wee free kings

skeptical scientist
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### Re: Analysis HW, compactness and uniform continuity.

I'm assuming the definition of compactness you were given is that every open cover has a finite subcover.

What do you know about sequences of points in compact sets? Qaanol suggests using "sequential compactness"—there's another equivalent* definition of compactness that talks about sequences.

*for metric spaces
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

polymer
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Joined: Mon Feb 04, 2008 7:14 am UTC

### Re: Analysis HW, compactness and uniform continuity.

Okay, I'm not sure Rudin plainly stated the criterion for sequential compactness. So I attempted to prove the criterion was true.
excuse me if the notation is a little non standard, I'm in a bit of a rush.

A subset E is compact in a metric space X if for any open cover of E, there exists a finite sub-cover of E.

Thm: E is compact in X iff for all sequences in E there exists some sub-sequence that converges to a limit in E.

(=>)

Assume E is compact in X, and fix some sequence p(n) in E. If the range of p(n) is finite, then p(n) is equal to a value in the range for infinitely many values, hence we can choose a sub-sequence that consists of this value, which converges to this value, which is in E. If the range is infinite, every infinite subset of a compact set has a limit point in the compact set, Hence there is a sub-sequence of p(n(k)) that converges to a point p in E.

OHMYGOODNESSTHISISWHATINEED

I'll try to prove the second half of this theorem later, because it's useful...

Continuing off of the hw problem:

The sequences get arbitrarily close as n goes to infinity. I can choose subsequences of those which approach a limit in E. Because the difference between them get arbitrarily small, they approach the same limit. Then the function difference being larger than epsilon contradicts continuity. This contradiction establishes the theorem in the HW. Thanks a bunch!

Quick question, I'm finding analysis to be much harder then I was expecting it to be, So I'm certainly going to be tempted to look for more help in the future. Is asking for help one or two times a week annoying?

skeptical scientist
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### Re: Analysis HW, compactness and uniform continuity.

polymer wrote:OHMYGOODNESSTHISISWHATINEED

That is not a word.
Continuing off of the hw problem:

This is almost correct, but you're missing something.

The sequences get arbitrarily close as n goes to infinity. I can choose subsequences of those which approach a limit in E.

Okay, so you can choose subsequences pk_n and qi_n that converge.
Because the difference between them get arbitrarily small, they approach the same limit.

You know d(pn,qn) -> 0 as n -> ∞, but how do you know d(pk_n,qi_n) goes to 0? (In fact, it might not; you need to be a bit more careful about how you choose your subsequences.)
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

polymer
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### Re: Analysis HW, compactness and uniform continuity.

Math is hard...

find a sub-sequence pk_n that converges to a limit in E, Then d(pk_n,qk_n) -> 0 as n -> ∞, and d(f(pk_n),f(qk_n)) > epsilon. But this contradicts the definition of continuity.

Thanks for the help.

skeptical scientist
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### Re: Analysis HW, compactness and uniform continuity.

polymer wrote:Math is hard...

find a sub-sequence pk_n that converges to a limit in E, Then d(pk_n,qk_n) -> 0 as n -> ∞, and d(f(pk_n),f(qk_n)) > epsilon. But this contradicts the definition of continuity.

That works. If you choose a subsequence of pn which converges, then the fact that d(pn,qn) -> 0 implies that the same indexed subsequence of qn also converges, and to the same limit.

There's a standard trick that would have worked as well, which is worth knowing. If you didn't know that d(pn,qn) -> 0, you could still do the following: let (pk_n) be a convergent subsequence of (pn), and then take (qi_n) to be a convergent subsequence of (qk_n). Since (pi_n) is a subsequence of (pk_n), which is convergent, you know that both (pi_n) and (qi_n) are convergent, so you get a single subsequence which converges for both. This works for any finite number of sequences, and with a little more work, can be made to work for a countably infinite collection of sequences.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

Qaanol
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### Re: Analysis HW, compactness and uniform continuity.

I’ve been trying to prove this without using sequential compactness, meaning everything in terms of open sets, and I think I succeeded.

Spoiler:
Let M, N be metric spaces. Let M be compact, meaning every open cover of M has a finite subcover. Let f:M→N be a continuous function, meaning for every open set in N, the preimage of that set under f is open in M.

Theorem: f is uniformly continuous, meaning ∀ε>0, ∃δ>0 such that dM(x, y)<δ ⇒ dN(f(x), f(y))<ε.

Proof: Given ε>0, consider all open balls of radius ε/2 with center in f(M)⊆N. That is, for all x∈M, consider the ball Bε/2(f(x)) = {z∈N : dN(f(x), z)<ε/2}. Note that these balls cover the image of M. Since these balls are open in N and f is continuous, the preimage f-1(Bε/2(f(x))) of each ball is open in M. For convenience, call that preimage U(x, ε) = f-1(Bε/2(f(x))), which is open in M.

For each U(x, ε), let Br(x)(x) be an open ball in M centered at x, with positive radius r(x), that is a subset of U(x, ε). There must be some such ball with positive radius at each point x, because U(x, ε) is open.

Let s(x) = ½r(x), so the set of balls Bs(x)(x) is an open cover of M. This has a finite subcover, because M is compact. Choose a finite subcover Bs1(x1) through Bsn(xn). Take δ = min(s1, …, sn).

Now consider any two points v, w ∈ M with dM(v, w) < δ. We know for some k that v∈Bsk(xk), so by the triangle inequality w is within sk+δ ≤ 2sk = rk of that same xk.

Thus both v and w are in Brk(xk) ⊂ U(xk, ε). Therefore f(v) and f(w) are both in Bε/2(f(xk)), and thus within ε of each other.
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Yakk
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### Re: Analysis HW, compactness and uniform continuity.

...

Style wise, I don't like how you introduced r(x). It is overly passive, and used before it is defined.

...

When talking about uniform continuity, my preference is to make the mapping between epsilon and delta explicit. Ie, define a function that, for a given epsilon, produces a delta. (Use [imath]\mu(\epsilon) = \delta[/imath], for example. Mu measures how "steep" your function is at a given level of resolution, in a sense. Then your proof reduces to finding such a mu, and proving that it has the property you want.

This reduces the degree of "passive voice" in the proof, and is logically equivalent.

Other than that, your proof looks fine to me.
skeptical scientist wrote:
polymer wrote:OHMYGOODNESSTHISISWHATINEED
That is not a word.
Acronym probably.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Marbas
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### Re: Analysis HW, compactness and uniform continuity.

Yakk wrote:
skeptical scientist wrote:
polymer wrote:OHMYGOODNESSTHISISWHATINEED
That is not a word.
Acronym probably.

Well, if you'll notice, there are twenty-seven letters in OHMYGOODNESSTHISISWHATINEED
That is one more than the number of letters in the alphabet
There is only one pope
The pope is a christian
Christians go to church
Churches are buildings
Buildings eventually collapse
The collapse of society is the apocalypse
Therefore, Polymer is clearly trying to cause the apocalypse. I can only assume OHMYGOODNESSTHISISWHATINEED is actually the true name of Satan.
Jahoclave wrote:Do you have any idea how much more fun the holocaust is with "Git er Done" as the catch phrase?

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