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Let $X$ be a Hausdorff space and let $A,B\subseteq X$ two compact subspaces which don't intersect. Show exist $U,V\subseteq X$ open which don't intersect s.t $A\subseteq U,B\subseteq V$.

I thought taking $A\ni a\neq b\in B$ and since $X$ is Hausdorff, $$\exists S_{a,i}\in N(a),S_{b,j}\in N(b):S_{a,i}\cap S_{b,j}=\emptyset$$ and taking $S_a=\bigcap_{a\in A,i\in I}S_{a,i},S_b=\bigcap_{b\in B,j\in J}S_b$ as my $U$ and $V$.

The only problem is I don't use the compactness. What am I missing? Is my solution correct?

user642796
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    you cannot just intersect them, it will not be an open set in general, e.g. $\cap (-1/n,1/n)$. – Daniel Valenzuela Sep 08 '14 at 12:16
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    If you take their union, then you can't guarantee that they won't intersect. You need compactness to find a finite set of open sets - and an intersection of a finite set of open sets is open – Mathmo123 Sep 08 '14 at 12:23

2 Answers2

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Let $b \in B$.

For each $a \in A$, we have $U_a \ni a, \ V_a \ni b$ open such that $U_a \cap V_a=\emptyset$. The $U_a$ form an open cover of $A$ so have a finite sub cover $U_{a_1}, \ldots, U_{a_k}$.

Let $$U_b = U_{a_1}\cup\cdots\cup U_{a_k}$$ which is open as it is a union of a finite number of open sets. Let $$V_b = V_{a_1}\cap\cdots\cap V_{a_k}$$

The $V_b$ form an open cover of $B$, and for each $b \in B$, $V_b \cap U_b = \emptyset$. Also note that for every $b$, $A \subset U_b$.

Can you use the compactness of $B$ to finish the proof?

man_in_green_shirt
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Mathmo123
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  • $V_a_k$ is chosen like $U_a_k$ is found? if not I still don't undertand how the compactness of B helps. –  Sep 08 '14 at 12:26
  • For each $a$, we found $V_a \ni b$ and $U_a \ni a$. So for a member of the finite sub cover $U_{a_j}$ we have a corresponding $V_{a_j}$ – Mathmo123 Sep 08 '14 at 12:27
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    To finish, you need to use the fact that the $V_b$ are an open cover of $B$ to find a finite sub cover and with corresponding covers of $A$, and remember that a finite intersection of open sets is open – Mathmo123 Sep 08 '14 at 12:29
  • Written like this, isn't $U_b$ a $\textit{union}$ of a finite number of open sets? – man_in_green_shirt Sep 17 '15 at 11:58
  • To finish it: the $V_{b}$ form a cover of $B$, so there exists a finite number of them such that $\bigcup V_{b_i}$ is a finite subcover of $B$. As this is a union of open sets, it's open. Next, take the intersection of the corresponding $U_{b_i}$. This is an intersection of finitely many open sets, and hence it's open. Each $U_{b_i}$ contains $A$, and hence so does their union. Furthermore, each $U_{b_i}$ is disjoint from $V_{b_i}$, and hence $\bigcup U_{b_i} \cap \bigcap V_{b_i}=\emptyset$. So $\bigcup U_{b_i}$ and $\bigcap V_{b_i}$ are the two open sets we're looking for. – man_in_green_shirt Sep 17 '15 at 12:24
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Hint: you need compactedness to choose a finite subcover of an open cover (which you choose s.t. it has properties that you like).

How is this going to help you? Well, finite intersections of open sets are open...

Daniel Valenzuela
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