# $X$ Hausdorff $\Leftrightarrow$ $\forall$ disjoint compacts $K_1,K_2:\exists$disjoint open sets $U,V$ including $K_1,K_2$ respectively [duplicate]

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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?

## marked as duplicate by Arnaud D., dantopa, JMP, tilper, Alex ProvostMay 30 '17 at 16:38

• 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. – J.C. Jan 23 '16 at 21:45
• 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
• 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
• If $K_1$ is empty let $U$ be the empty set and $V=X,$ the whole space. – DanielWainfleet Jan 23 '16 at 22:57

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.
• 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. – J.C. Jan 23 '16 at 21:52