Compact sets in Hausdorff space are closed

[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.

Thanks in advance...

 

[Office_Shredder]

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

 

[SonyAD]

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...

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

 

[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.

 

[Fredrik]

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...

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.