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Show that a topological space $X$ is separated (Hausdorff) if and only if for all compact subspaces $K_1$ and $K_2$ of $X$, there are disjoint sets $U$ and $V$ of $X$ such that $K_1 \subseteq U$ and $K_2 \subseteq V$.

I have one way ($\Leftarrow$), but I'm having some problems finding a neat proof from left to right. I think I've found a solution, but it's quite tedious. Here it is:

We assume $X$ is separated and we take two disjoint compact subspaces $K_1$ and $K_2$ of $X$.

Let's fix $x \in K_1$. Then $\forall y \in K_2$, there are two disjoint open sets $V_{x,y}$ and $U_{x,y}$ including $x$ and $y$ respectively. All these $U_{x,y}$ cover $K_2$ and therefore we can extract a finite number of them that cover it too $\{U_{x,y_i} : i = 1,...,n\}$ ($n \in \mathbb{N}^{\ast}$), and we take the correspondent $\{V_{x,y_i} : i = 1,...,n\}$. Then the sets $V_x:= \bigcap_{i = 1}^n V_{x,y_i}$ and $U_x := \bigcup_{i = 1}^n U_{x,y_i}$ are open disjoint sets.

Now I do this for all $x \in K_1$ and get a set $\{V_x : x \in K_1\}$ of open sets covering $K_1$, and corresponding $\{U_x : x \in K_1\}$ such that $\forall x \in K_1, V_x \cap U_x = \emptyset$. Since $K_1$ is compact, I now extract a finite number of those open sets $\{V_{x_i}: i=1,...,m\}$ ($m \in \mathbb{N}$), which still cover $K_1$ and the corresponding $\{U_{x_i}: i=1,...,m\}$.

Now $V :=\bigcup_{i = 1}^m V_{x_i}$ is open and covers $K_1$. While $U :=\bigcap_{i = 1}^m U_{x_i}$ is open too and covers $K_2$. Moreover $U \cap V = \emptyset$. That finishes the proof.

Is this approach correct? But more importantly, isn't there a less tedious way?

Daniel Fischer
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K.A.
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    At the end, you probably meant $$\bigcap_{i = 1}^m U_{x_i}.$$ If you take the union, you can't guarantee disjointness. – Daniel Fischer Jan 23 '16 at 21:43
  • Yes of course, thank you. I meant the intersection, precisely because it guarantees me that the sets will be disjoint. – K.A. Jan 23 '16 at 21:45
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    In that case, your argument is correct, it's the standard argument (or at least one of the standard arguments), and I don't find it tedious. – Daniel Fischer Jan 23 '16 at 21:47
  • I have to admit that my initialy proof (when I to write it to post here) was way more tedious, but then I remembered that if I kept those $U_{x_i}$ sets, that I can skip a lot of work. I still posted it, because the $\Leftarrow$ way of the proof is like two lines long, so I wondered if there is a faster way for $\Rightarrow$ too – K.A. Jan 23 '16 at 21:49
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    Well, one of the conditions is rather immediately stronger than the other. So for the one direction, you only need to observe that singleton sets are compact. It's not surprising that proving the prima facie stronger condition from the prima facie weaker one is more complicated. – Daniel Fischer Jan 23 '16 at 21:54
  • Nice, thank you. This also confirms my idea on the other direction. $x \neq y$, then ${x}$ and ${y}$ are compact, so there are disjoint open sets around them. When things are too easy, I'm always a bit suspicious. – K.A. Jan 23 '16 at 21:56
  • This type of 2-stage proof occurs frequently in topology. Minor point :You left out the trivial case where $K_1$ or $K_2$ is empty. – DanielWainfleet Jan 23 '16 at 22:38
  • Right, thank you. It seems like minor point, but it's actually important for me. – K.A. Jan 23 '16 at 22:39
  • @user254665: but what happens exactly when $K_1$ is empty? I can't really look for a neighbourhood around it. Or do I simply take $X$ as the neighbourhood of $K_2$ and $\emptyset$ as the one of $K_1$. Because those sets would be disjoint and indeed open, and contain $K_1$ and $K_2$ respectively. – K.A. Jan 23 '16 at 22:41
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    If $K_1$ is empty let $U$ be the empty set and $V=X,$ the whole space. – DanielWainfleet Jan 23 '16 at 22:57
  • How many times is this going to show up in the review queue? –  May 30 '17 at 13:27

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It's almost correct:

You have the $U_x$ and $V_x$ such that $x \in V_x$ and $K_2 \subseteq U_x$ and $U_x \cap V_x = \emptyset$.

You then have $V_x$ for every $x \in K$ and have finitely many that cover $K_1$. You are correct in using the union of the $V_x$ to cover $K_1$, but you need the intersection (!) of the corresponding $U_x$, which being a finite intersection, is still an open neighbourhood of $K_2$. The union won't do. Check the details: suppose $x$ is in the intersection of $U$ and $V$, where $U$ is defined as the intersection, then $x$ is in some $V_{x_i}$ but the intersection garantuees it is also in the same-indexed $U_{x_i}$ as well, getting an immediate contradiction. Otherwise we'd only know $x$ is in some $V_{x_j}$ and we'd know nothing. It works the same in the firts case, that you did get correct. It's directly analogous.

This is the standard way. Often the point vs compact case is a sublemma proved separately (namely that in a Hausdorff space we can separate a point and a compact set). I wouldn't know of any essentially simpler way.

Henno Brandsma
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  • Yes I know, someone already pointed it out. But thank you anyway. I wanted to use the intersection, but I copy-pasted the union and only change the $V$ into $U$, but forgot to change the "\bigcup" into "\bigcap".

    Thank you for confirming its correctness. I tend to prove things very tediously, and the $\Leftarrow$ proof was really fast, so I thought I was missing a detail that would make the proof much faster.

     – K.A. Jan 23 '16 at 21:52
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    @J.C. The meat is in this direction. Here we use the power of compactness to transfer facts about points to facts about compact sets. I don't find it tedious, it's almost the only thing one can do. – Henno Brandsma Jan 23 '16 at 21:54