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Near the top of page 25 of Steen & Seebach's "Counterexamples in Topology" (2nd edition) it says that if A is a compact subset of a T3 space, and U is an open set containing A, then there is an open set V such that $A\subset V\subset\bar{V}\subset U$ (where $\bar{V}$ is the closure of V). I can see that we can get an open set V from the intersection of U with the finite open cover of A, but I've been banging my head against the wall trying to figure out why $\bar{V}\subset U$. Can anyone enlighten me, or it is a mistake in the book?

Ben
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  • $T_3$ means the closed neighbourhoods form a neighbourhood basis at each point. – Daniel Fischer Nov 07 '20 at 18:28
  • @Daniel Fischer: how does that help? A could contain infinitely (or uncountably) many points. – Ben Nov 07 '20 at 18:35
  • But $A$ is compact. Thus finitely many neighbourhoods suffice. – Daniel Fischer Nov 07 '20 at 18:36
  • @Daniel Fischer: in S&S's book, T3 is defined as separation of points from closed sets by disjoint open sets. Where do you get the idea/definition(?) about closed neighbourhoods forming a local basis? – Ben Nov 07 '20 at 19:05
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    It's a useful equivalent criterion. (Point $x$, open neighbourhood $U$; let closed set $F = X \setminus U$. Separate $x$ and $F$ by $V$ and $W$, then $\overline{V} \subset U$. Conversely, if $F$ is closed and $x \notin F$, then $U = X\setminus F$ is a neighbourhood (open) of $x$. There is a closed neighbourhood $V$ of $x$ contained in $U$, then $X\setminus V$ and $V^{\circ}$ are open sets separating $x$ and $F$.) – Daniel Fischer Nov 07 '20 at 19:11

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It is easy to show this to be true when $A$ is a point, so in general there is a cover $\{V_x\}_{x\in A}$ of $A$ such that each $\bar V_x$ is contained in $U$. By compactness of $A$ there is a finite subcover $\{V_i\}_1^n$. Set $V = \cup_1^n V_i$, then $$A\subset V \subset \bar V = \cup_1^n \bar V_i \subset U.$$

abhi01nat
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  • I just reached the same conclusion after looking at this post: https://math.stackexchange.com/questions/18192/disjoint-compact-sets-in-a-hausdorff-space-can-be-separated Is it true to say that the closure of the union is the same as the union of the closure though? i.e. shouldn't the equality symbol be replaced by a subset symbol? – Ben Nov 07 '20 at 18:49
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    @Ben We have $\overline{V_i} \subseteq \overline{V}$ for all $i$, hence equality. – Daniel Fischer Nov 07 '20 at 18:59