Suppose that $U$ is an open subset of a topological space $S$ and that $p\in U$ is a point of $U$. Does there exist a closed subset $A\subset S$ such that $p\in A \subset U$ ?!. Is even this statement true for certain topological spaces ?.
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Did you try something? What you asked, although not true in general, holds for basically any topological space that you can think of. – José Carlos Santos Jul 10 '17 at 13:49
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This is certainly true in spaces where points are closed -- for instance, metric spaces. (More generally, in $T_1$ spaces. The $T_1$ condition is actually equivalent to all singleton sets being closed.)
You can come up with counterexamples, though. For a somewhat contrived example, take the set $\{A,B\}$ with open sets $\emptyset$, $\{A\}$, $\{A,B\}$. Then take $U=\{A\}$, $x=A,$ and you'll see the open set $U$ contains no closed set containing $x$.
vociferous_rutabaga
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Can we choose a closed subset $B$ about $x$ such that $B$ contains an open subset $O$ containing $x$ (i.e., $x\in O \subset B$)? – Hussein Eid Jul 10 '17 at 14:01
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The entire topological space is open and closed, so this trivially gives such a $B$ (with $B=O$ for any $x$). Or, take any open set containing $x$ and take $B$ its closure, and you'll have $x \in O \subset B$. (For some $x$ the only possible $O$ might be the entire space, in which case this second suggestion will recover the first.) – vociferous_rutabaga Jul 10 '17 at 14:15
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This is not true for all topological spaces.
Consider for example the set $\{a, b, c\}$ with the open sets $\{a, b, c\}, \{a, b\}, \{b, c\}, \{b\}$ and $\emptyset$.
This forms a topological space but there is no closed set contained in $\{b\}$ that contains the point $b$, because the complement $\{a, c\}$ is not open.
Tob Ernack
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