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Claim:If $A,B$ are compact disjoint subsets of the Hausdorff space $X$, then there exists disjoint open sets $U,V$ containing $A,B$ resp.

Would I be on the right track in saying that since $A,B$ are compact subsets of $X$ then choose $\{\mathbb{A}_\alpha \}$, $\{\mathbb{B}_ j \}$ to be open covers for $A,B$ resp. Then since $X$ is Hausdorff we have that for each $x$ in $\{\mathbb{A}_\alpha \}$ and $y \in \{\mathbb{B}_j \}$ there exists disjoint open sets $U,V$.

Now if we take $U = \cup_{x \in \mathbb{A}_\alpha} U_x$ and $V = \cup_{y \in \mathbb{B}_j} V_y$ we have our disjoint open sets.

Comment: However, I didn't uses the fact that $A,B$ were disjoint, or that they are closed, since every compact subset of a Hausdorff space is closed.

Can someone give me some useful hint? This is hw, so I don't want and answer. Thanks in advance.


marked as duplicate by Arnaud D., Claude Leibovici, JonMark Perry, dantopa, tilper May 30 '17 at 13:24

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I see no reason to think your $U$ and $V$ are disjoint. The $U_x$ and $V_y$ (for some pair $(x \in A_\alpha, y \in B_j)$ could intersect the $V_z$ and $U_w$ for some other pair $(w \in A_{\beta}, z \in B_k)$.

Hint: start by saying that for each $x \in A$ and $y \in B$ there are disjoint open sets $U_{x,y}, V_{x,y}$ with $x \in U_{x,y}$, $y \in V_{x,y}$. The $V_{x,y}$ (for any given $x$) form an open cover of $B$ ...


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