How do I turn my verbal argument into something formal in [Real Analysis]? (proving every compact set is bounded)

Question

So one of the exercises I am doing is to prove (or disprove) that 'Every compact set on a metric space is bounded'. Verbally, I can 'prove' this by simply stating: "If the every compact set on a metric space is not bounded, then there exists an infinite number of open covers, and if something is compact, there are only a finite number of open covers. This is a contradiction, and thus every compact set on a metric space is bounded."

Two questions: is this argument right? And if so, how to I learn how to represent this formally?

Answer 1

Your argument is essentially correct but pay attention to the essentially.

I would phrase it in words like so:

If not every compact set on a metric space is bounded then there exists at least one compact set on a metric space that is not bounded. Since it's unbounded we can construct an infinite open covering which doesn't have a finite subcover thus this unbounded compact set cannot actually exist.

Note the few differences I have

• First I say there exists at least one compact set on a metric space that is not bounded. This is needed because we need some sort of compact set to work with - otherwise we're really not being precise with what we're saying since every compact set on a metric space is not bounded could be misleading (one could think you're stating every compact set on a metric space is unbounded, which is subtly different).

• Next I say Since it's unbounded we can construct an infinite open covering which doesn't have a finite subcover. This is the very important part and why I said "pay attention to the essentially". The root of the contradiction is that there does exist an open cover without a finite subcover, which is exactly what I stated here.

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Now I encourage you to make things mathy, but below is what I came up with.

Claim: Let $(X,d)$ be a metric space, then every compact subset of $X$ is bounded. Proof: Assume $\exists A \subset X$ compact and unbounded. Put $\mathcal{V} = \{ B_0(n) \mid n \in \mathbb{N} \}$ then $\mathcal{V}$ is an open covering of $A$ but there doesn't exist a finite subcover, since if there was one then $A$ would become bounded. Note that $B_0(n)$ is the ball centered at $0$ of radius $n$.

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I want to add the following comment: When I said your argument is essentially correct I meant that it appears you had the essence of the proof in mind, not that you could use this argument exactly how it is for a precise, formal or really even informal proof. I apologize for being misleading - from my perspective it seems like you have an idea on what's going on but are unfamiliar with wording things in a precise manner (in some sense this is exactly what math is), but I would still argue that having this base intuition is also important. To put it concisely, sorry for being misleading, you're argument is not valid, but it appears (at least to me) that you have the proper intuition.

Answer 2

I agree with J. Loreaux's commrent above (“Your argument doesn't really work”), and I'd go farther: to me, your argument makes no sense whatever, for several reasons:

1. You say “If the every compact set on a metric space is not bounded, then …”. This is at least confusingly stated. You want to prove that every compact set is bounded. You seem to be trying to do a proof by contradiction. To do this, you need to assume the negation of the statement “Every compact set is bounded”. The negation of this statement is “there exists an unbounded compact set”. But instead you seem to be assuming that every compact set is unbounded. This is a much stronger assumption, and proving a contradiction from it will not allow you to conclude that every compact set is bounded; it will only allow you to conclude that some compact set is bounded, which is useless. (There are much simpler arguments to prove that.)

   Note that the statements “every cat is not black” and “not every cat is black” are quite different. The first is false, but the second is true.

2. Your conclusion from the puzzling assumption is that “there exists an infinite number of open covers,” which is puzzling because an open cover has to cover something and you didn't say what was being covered. In the context of a particular set $S$ which might or might not be compact, the reader will assume from context that the phrase “open cover” means “open cover of $S$”. But there is no such context here because the preceding phrase is talking about every compact set.

3. “if something is compact, there are only a finite number of open covers” is strange and, at least as you stated it, false. There are any number of examples of compact sets which have an infinite family of open covers.

I could go on, but I think this is enough. I think many people reading your argument, including the grader in your class, will conclude that you have no idea what a compact set is. Perhaps they would be wrong, but if so you have to do better or that is what they are likely to think.

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An open cover $C$ of some set $S$ is a family of open sets whose union contains $S$.

A set $S$ is compact if every open cover of $S$ has a finite subcover. That means that if you have a family $C$ of open sets whose union contains $S$, you can always find a finite subset $C'\subset C$ whose union also contains $S$.

Here we want to show that if $S$ is a compact subset of a metric space, then $S$ is bounded. Take an arbitrary compact set $S$. If we can show that $S$ is bounded, we are done. Find a cover $C$ of $S$ whose members are open balls of radius 1. (You need to fill in the details to show that such a cover $C$ of $S$ exists.) Let $C'$ be finite subcover of this set $C$ of open balls which still covers $S$; we can find $C'$ because $S$ is compact. We now have a covering of $S$ with a finite set of open balls of radius 1. But we can then use the triangle inequality to show that $S$ has bounded diameter. (You need to fill in some details here too.)

(I see that the argument of DanZimm elsewhere in this thread is somewhat simpler.)