Compact sets in Hausdorff space are closed

1. Jun 25, 2010

mordechai9

First of all I just want to rant why is the Latex preview feature such a complete failure in Firefox? Actually it is really bad and buggy in IE too....

So I am reading into Foundations of geometry by Abraham and Marsden and there is a basic topology proof that's giving me some trouble. They define compact spaces as saying a space S is compact iff every open covering $$S = \cup U_\alpha$$ has a finite subcovering, and if A is a subset of S, then any open cover has a finite subcover in the relative (subspace) topology. I am pointing out this definition just because in other places I think I've seen open covers defined as $$S \subset \cup U_\alpha$$ , not necessarily $$S = \cup U_\alpha$$.

Then they propose that in a Hausdorff space, all the compact subsets are closed. For proof, let A be a compact subset of S. Then let $$u \in A^C$$ and $$v \in A$$ have open disjoint neighborhoods $$U_u, U_v$$. Since A is compact we can write

$$A = \cup_{v \in A} U_v \cap A$$

for a finite number of $$v \in A$$. Furthermore each of these neighborhoods must be disjoint from $$U_u$$. Then they conclude there are disjoint neighborhoods for u and A , and $$U_u \subset A^c$$, so A^c is open.

I don't understand this. First of all, it seems the cover for A isn't a neighborhood, since it is only open in the relative topology, but maybe that's what they mean to say. In other words, maybe they mean A has a neighborhood in the relative topology.

More importantly, it doesn't seem like we needed to use the compactness here. We have an open neighborhood for u, and we need to show that neighborhood is entirely contained in $$A^c$$, so we need to show

$$U_u \cap A = \oslash$$,

and we have

$$U_u \cap A = U_u \cap ( \cup U_v \cap A ) = \oslash$$.

That follows without compactness, but obviously I know that can't be right. If that were true, all subsets would be closed, which would be ridiculous.

I figure the problem is that we have to worry if $$U_u \cap U_v \ne \oslash$$ for some $$v \in A$$, in which case, we use the compactness of A to exempt that $$U_v$$ from the cover. But it seems like we still don't know that the neighborhoods we have left will be able to cover all of A.

2. Jun 25, 2010

Office_Shredder

Staff Emeritus
You can pick $$U_v$$ and $$U_u$$ to be open in the Hausdorff space, and then the $$U_v$$'s will be an open cover of A because their intersection with A is open in the induced topology (by definition)

The key now is the finiteness. We don't have a single $$U_u$$ for every $$U_v$$, we have to pick a different one for every choice of v. So maybe it would be better to denote the sets around $$u$$ as $$V_v$$ (sorry for the confusing notation but I want to change as little as possible). Then to find the open neighborhood around $$u$$ we need to intersect every $$V_v$$ to get an open set that is disjoint from every $$U_v$$ simultaneously. Intersections of open sets are only open in general if there are finitely many of them, which is where compactness comes into play: we only needed to look at finitely many v's

3. Jun 25, 2010

It isn't a failure. It just uses image caching. You need to manually refresh.

4. Jun 26, 2010

mordechai9

Ah, ok, I see. That's more or less what I was thinking I just wasn't sure how to express the details correctly. I also get how the U_v's form an open cover in the induced topology, I just thought that maybe they were saying they were also open in the general topology. Thanks for the the help.

5. Jun 26, 2010

Fredrik

Staff Emeritus
It's pretty much the same regardless of what browser and operating system you're using. Been that way for a couple of months I think. No one seems to know what's causing it or how to fix it. The workaround is to refresh and resend (SonyAD didn't mention the resend). Works with all browsers and operating systems.