-1
$\begingroup$

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 ?.

$\endgroup$

closed as off-topic by José Carlos Santos, Sahiba Arora, B. Goddard, Davide Giraudo, Harambe Jul 10 '17 at 23:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Sahiba Arora, B. Goddard, Davide Giraudo, Harambe
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Did you try something? What you asked, although not true in general, holds for basically any topological space that you can think of. $\endgroup$ – José Carlos Santos Jul 10 '17 at 13:49
1
$\begingroup$

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 $x$.

$\endgroup$
  • $\begingroup$ Can we choose a closed subset $B$ about $x$ such that $B$ contains an open subset $O$ containing $x$ (i.e., $x\in O \subset B$)? $\endgroup$ – Hussein Eid Jul 10 '17 at 14:01
  • $\begingroup$ 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.) $\endgroup$ – vociferous_rutabaga Jul 10 '17 at 14:15
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.