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I was given a claim:

Let $(X, \mathfrak{T})$ be a topological space. Then $X$ is a regular space iff $\forall x \in X, \forall U \in \mathfrak{T}$ s.t. $x \in U$, $\exists V$ such that $x \in V \subseteq \overline V \subseteq U$

Regular is defined as (similar to wiki):

If, given any point $x$ and closed set $F$ in $X$ such that $x$ does not belong to $F$, they are separated by open disjoint neighbourhoods

I was stuck on a minor part of the proof and need some help!


$(\Rightarrow)$ Let $(X, \mathfrak{T})$ be a regular space. Then $\forall x \in X$, $F \subset X$, $F$ is closed and does not contain $x$, there exists open sets $U$ and $W$, such that $x \in U, F \subset W$ and $U \cap W = \varnothing$.

Since $U$ is open, for all points $X$, there exists an open set $V$ such that $x \in V \subset U$. (We have just produced $V$ in the claim)

Let $y \in F \subset W$ be another point. Then by definition of regular, $y$ is separated from any closed set. Consider the closed set $\overline V$, then we have $y \in W, x \in V \subseteq \overline V \subseteq U$. And since $U \cap W = \varnothing, y \notin \overline V$

($\qed$ $\eop$ <--- how to generate end of proof symbol?)


There is something terribly wrong in the second last line of the proof which I stated:

" $x \in V \subseteq \overline V \subseteq U$"

There is no insurance that such $\overline V \subseteq U$. When is it true that $V \subseteq \overline V \subseteq U$ will hold for open sets? It is very likely that $U \subseteq \overline V$ instead.

How do I correct this line?

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  • $\begingroup$ Your choice of using $\subseteq$ only once in the statement is strange. It's likely is not true if $\subset$ indicates proper inclusion. $\endgroup$ – Matt Samuel Jun 11 '16 at 21:03
  • $\begingroup$ @MattSamuel Sorry typo $\endgroup$ – Shamisen Expert Jun 11 '16 at 21:06
  • $\begingroup$ end of proof: use \box or \blacksquare $\endgroup$ – ArtW Jun 11 '16 at 21:06
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You seem to want to prove that, for a topological space $X$, the following conditions are equivalent:

  1. for all $x\in X$ and all open subsets $U$ such that $x\in U$, there exists an open set $V$ such that $x\in V$ and $V\subseteq \bar{V}\subseteq U$

  2. for all $x\in X$ and all closed subsets $F$ such that $x\notin F$, there are disjoint open sets $U$ and $V$ such that $x\in V$, $F\subseteq U$

Proof of 1$\implies$2

Let $x\in X$ and let $F$ be a closed subset with $x\notin F$. Then $W=X\setminus F$ is an open set and $x\in W$; by hypothesis, there exists an open set $V$ with $x\in V$ and $V\subseteq\bar{V}\subseteq W$. Then $U=X\setminus\bar{V}$ is an open set and $V\cap U=\emptyset$.

Proof of 2$\implies$1

Let $x\in X$ and let $U$ be an open set with $x\in U$. Then $F=X\setminus U$ is a closed set and $x\notin F$. By assumption, there are disjoint open sets $V$ and $W$ such that $x\in V$ and $F\subseteq W$. The second condition means $U=X\setminus F\supseteq X\setminus W=G$. Since $V\cap W=\emptyset$, we have $V\subseteq G$; therefore $\bar{V}\subseteq G$. Thus $V\subseteq\bar{V}\subseteq U$.


You're not choosing carefully the sets, this is the source of your problems.

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  • $\begingroup$ In proof 2 $\implies$ 1 how did you know that since $V \subseteq G$, therefore $\overline V \subseteq G$ as well? $\endgroup$ – Shamisen Expert Jun 12 '16 at 2:24
  • $\begingroup$ @TheSilenceoftheCows Because $G$ is closed (complement of open set $W$), and so $V \subseteq G$ implies $\overline{V} \subseteq \overline{G} = G$. $\endgroup$ – Henno Brandsma Jun 12 '16 at 6:03
  • $\begingroup$ @HennoBrandsma I don't know this result, but I can conceptualize it as: since any sequence in $V$ is a sequence in $G$, therefore all the limits of sequences in $V$ are in $G$. The set of all limits of sequences in $V$ is $\overline V$, all the limits are in $G$, therefore $\overline V \subseteq G$. Is there a more topological way of thinking about this containment relationship? i.e. any open in closed set $C$ implies closure of open is in closed set $C$ $\endgroup$ – Shamisen Expert Jun 12 '16 at 6:40
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    $\begingroup$ Don't think in terms of sequences (only works in some spaces). Another way to see this: the closure of $V$ is the smallest closed set containing $V$, and $G$ is one of those closed sets continaing $V$, so $\overline{V} \subseteq G$ by the minimality. $\endgroup$ – Henno Brandsma Jun 12 '16 at 8:31
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Here's an outline for a proof of your claim assuming $X$ is regular.

Let $x\in X$ be any point and let $U$ be some open set containing $x$. We know that $X\setminus U$ is closed so by regularity there exist disjoint open sets $V,W$ with $x\in V$ and $X\setminus U\subseteq W$. Let $y$ be a limit point of $V$. The point $y$ is not in $W$, because otherwise we would find an open set $U'$ such that $y\in U'\subseteq W$, but this contradicts that $V\cap W=\emptyset$ since $U'$ must contain points of $V$. Hence $y\not\in W$ which implies $y\in U$ and it follows $\overline{V}\subseteq U$.

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