- #1
mordechai9
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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 [tex] S = \cup U_\alpha [/tex] 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 [tex] S \subset \cup U_\alpha [/tex] , not necessarily [tex] S = \cup U_\alpha [/tex].
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 [tex] u \in A^C [/tex] and [tex] v \in A [/tex] have open disjoint neighborhoods [tex] U_u, U_v [/tex]. Since A is compact we can write
[tex] A = \cup_{v \in A} U_v \cap A [/tex]
for a finite number of [tex] v \in A [/tex]. Furthermore each of these neighborhoods must be disjoint from [tex] U_u [/tex]. Then they conclude there are disjoint neighborhoods for u and A , and [tex] U_u \subset A^c [/tex], 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 [tex] A^c [/tex], so we need to show
[tex] U_u \cap A = \oslash [/tex],
and we have
[tex] U_u \cap A = U_u \cap ( \cup U_v \cap A ) = \oslash [/tex].
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 [tex] U_u \cap U_v \ne \oslash [/tex] for some [tex] v \in A [/tex], in which case, we use the compactness of A to exempt that [tex] U_v [/tex] 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...
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 [tex] S = \cup U_\alpha [/tex] 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 [tex] S \subset \cup U_\alpha [/tex] , not necessarily [tex] S = \cup U_\alpha [/tex].
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 [tex] u \in A^C [/tex] and [tex] v \in A [/tex] have open disjoint neighborhoods [tex] U_u, U_v [/tex]. Since A is compact we can write
[tex] A = \cup_{v \in A} U_v \cap A [/tex]
for a finite number of [tex] v \in A [/tex]. Furthermore each of these neighborhoods must be disjoint from [tex] U_u [/tex]. Then they conclude there are disjoint neighborhoods for u and A , and [tex] U_u \subset A^c [/tex], 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 [tex] A^c [/tex], so we need to show
[tex] U_u \cap A = \oslash [/tex],
and we have
[tex] U_u \cap A = U_u \cap ( \cup U_v \cap A ) = \oslash [/tex].
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 [tex] U_u \cap U_v \ne \oslash [/tex] for some [tex] v \in A [/tex], in which case, we use the compactness of A to exempt that [tex] U_v [/tex] 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...