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Suppose that $X$ is a compact Hausdorff space and let $\{x_1,\ldots,x_n\}$ be a finite collection of points in $X$. It it possible to find open neighbourhoods $U_i$ of $x_i$ such that such that $x_j \notin \overline{U_i}$ for all $j \ne i$ and $\overline{U_i} \cap \overline{U_j} = \varnothing$ for all $i \ne j$? If not what extra separation axioms are needed?

I know for example that $X$ being compact Hausdorff implies that $X$ is normal, but that doesn't seem strong enough.

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Hint: use the fact that compact Hausdorff spaces are normal and regular.

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Thanks to Kavi for the hint! It made me realise what I was missing. Here's a rough outline of the answer for completion.

Use regularity of $X$ to obtain disjoint open sets $U$ and $V$ such that $x_1 \in U$ and $\{x_2, \ldots, x_n\} \subseteq V$. Then $U \subseteq V^c$ so that $\overline{U} \subseteq V^c$. In particular, $\overline{U} \cap V = \varnothing$. Now use normality of $X$ to obtain disjoint open neighbourhoods $W$ of $\overline{U}$ and $Y$ of $\{x_2, \ldots, x_n\}$. Then $Y \subseteq W^c$ so that $\overline Y \subseteq W^c \subseteq \overline{U}^c$. Hence, $\overline{Y} \cap \overline{U} = \varnothing$. A similar process can be undertaken to separate off the other points.

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  • $\begingroup$ normality is overkill. $\endgroup$ – Henno Brandsma Jan 22 at 6:38
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You can use Hausdorffness to separate the $x_i$ into pairwise disjoint $U_i$ such that $x_i \in U_i$, $i=1,\ldots,n$ (this follows by a simple induction on $n$ that this can be done in any Hausdorff space) and then regularity to find $x_i \in V_i \subseteq U_i$, $V_i$ open, with $\overline{V_i} \subseteq U_i$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.

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