Compact Sets

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Show that every compact set must be closed.

I am looking for a simple proof.

This is supposed to be Intro Analysis proof.

Shadow · **Posted:** Tue, 4 Oct 2011 00:15:06 UTC

glebovg wrote:

Show that every compact set must be closed.

I am looking for a simple proof.

This is supposed to be Intro Analysis proof.

Are you in a metric space?

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

Unfortunately, we have not covered metric spaces, yet. I am supposed to use covers, open covers, subcovers, etc. and the fact that any compact set must be bounded.

Shadow · **Posted:** Tue, 4 Oct 2011 00:34:41 UTC

glebovg wrote:

Unfortunately, we have not covered metric spaces, yet. I am supposed to use covers, open covers, subcovers, etc. and the fact that any compact set must be bounded.

Well, you'll at least need that your ambient space is Hausdorff. Then let

be compact. Select a point in the complement of

and a point in

and neighborhoods around each which do not intersect. Continue to produce a cover for

, and extract a finite subcover. Then you can argue that the intersection of the finitely many complement neighborhoods does not intersect

, and so points in the complement have neighborhoods around them, hence the complement is open.

Shadow · **Posted:** Tue, 4 Oct 2011 00:35:13 UTC

Also, this belongs in Analysis & Topology. Topic moved.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

How would I argue that the intersection of the finitely many complement neighborhoods does not intersect K?

Shadow · **Posted:** Wed, 5 Oct 2011 17:14:23 UTC

glebovg wrote:

How would I argue that the intersection of the finitely many complement neighborhoods does not intersect K?

You must have a Hausdorff space for this.

Actually, your questions all seem to have to do with $\mathbb{R}$ , so far. . . is your ambient space supposed to be $\mathbb{R}$ ? (i.e. are you assuming $K\subsetq\mathbb{R}$ ? If so, then you REALLY should have stated that at the beginning, because there is a lot of extra information in that that one could be using. If not, then you need the space to be assumed Hausdorff so you can separate points like that.

outermeasure · **Posted:** Wed, 5 Oct 2011 17:34:51 UTC

Shadow wrote:

glebovg wrote:

How would I argue that the intersection of the finitely many complement neighborhoods does not intersect K?

You must have a Hausdorff space for this.

Actually, your questions all seem to have to do with $\mathbb{R}$ , so far. . . is your ambient space supposed to be $\mathbb{R}$ ? (i.e. are you assuming $K\subseteq\mathbb{R}$ ? If so, then you REALLY should have stated that at the beginning, because there is a lot of extra information in that that one could be using. If not, then you need the space to be assumed Hausdorff so you can separate points like that.

Hausdorff is sufficient, but not necessary. In fact, the property "compact=>closed" lies strictly between T_1

(singletons are closed) and T_2

(Hausdorff) --- the cofinite topology on an infinite set is a T_1

space where compact does not imply closed (since every subset is compact), and cocountable topology on an uncountable set is an example of "compact=>closed" (since the only compact sets are the finite sets) which is not T_2

.

glebovg · **Joined:** Mon, 28 Dec 2009 00:16:28 UTC **Posts:** 228

This is Intro Analysis proof. Yes, K subset R. I am trying to prove it using basic Intro Analysis techniques. I presume that theorems like this one were proven before Topology became a major area of mathematics. Does it not seem like cheating (proving Analysis theorems using techniques from Topology given that these theorems were proven without using techniques from Topology)?

Shadow · **Posted:** Wed, 5 Oct 2011 20:59:14 UTC

glebovg wrote:

This is Intro Analysis proof. Yes, K subset R. I am trying to prove it using basic Intro Analysis techniques. I presume that theorems like this one were proven before Topology became a major area of mathematics. Does it not seem like cheating (proving Analysis theorems using techniques from Topology given that these theorems were proven without using techniques from Topology)?

Not really, Hausdorff is just a property that spaces have, it's not really cheating. Someone proves something in like $\mathbb{R}$ then notices that they didn't really CARE that they were inside the real numbers, all they cared about was that they were able to separate two points by open sets, and defined Hausdorff spaces to be the places where you can do that.

In other news: PLEASE include ALL assumptions in the future, this could have been vastly simplified if we had known that $K\subseteq\mathbb{R}$ from the start rather than just an abstract compact set in some ambient space.

My post which starts with "Well, you'll at least need that your ambient space is Hausdorff[. . .]" has what you need, you'll need to show that you can always separate two points in $\mathbb{R}$ by neighborhoods, but that is easy, just choose $a\ne b\in\mathbb{R}$ and let $\varepsilon={|a-b|\over 2}$ and then the $\varepsilon$ -neighborhood of

doesn't intersect the $\varepsilon$ neighborhood of

, then proceed with the rest of the proof that the complement is open.

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