6
$\begingroup$

Let $X$ be a topological space (no assumptions about separation), $U$ be a regular open subset of $X$ (i.e. $\mbox{int}\,\mbox{cl}\, U = U$) and $D$ a dense subset of $X$. Is $D \cap U$ a regular open of $D$, i.e. is $\mbox{int}_D \mbox{cl}_D (D \cap U) = D \cap U$?

$\endgroup$

1 Answer 1

1
$\begingroup$

I do not know if in some point I make some mistake. $U\cap D$ should be open in the induced topology over $D$ (this is due to the density of $D$ inside $X$).

Suppose now that $\exists x\in\mbox{int}_D\mbox{cl}_D (U\cap D) \setminus (U\cap D) $. This means:

  1. $x\notin U$
  2. $x\in \mbox{cl}_D(U\cap D) $
  3. $\exists \bar{U}^{(D)}_x$ open neighbourhood of $x$ in $D$ such that $ \bar{U}^{(D)}_x\subseteq \mbox{cl}_D(U\cap D)$

From 2 we have that $\forall U^{(D)}_x$, $U^{(D)}_x\cap D\cap U \neq \emptyset$ and in particular we have $U^{(D)}_x\cap U\neq \emptyset$. Since $D$ is dense it follow that $\forall U^{(X)}_x$ open in $X$ we have $U^{(X)}_x\cap U \neq \emptyset$.

Since $U$ is regular in $X$ and since $x$ is a point in the frontier of $U$, it follows that $\nexists U^{(X)}_x$ such that $U^{(X)}_x\subseteq \mbox{cl}_X(U)$ and you can see that this implies $\nexists\ U^{(X)}_x\cap D$ (or $U^{(D)}_x$) such that $U^{(D)}_x\subseteq \mbox{cl}_D(U)$. This contradicts point 3.

This final statement follows from two facts.

First we have that $\mbox{cl}_D(U)\subseteq \mbox{cl}_X(U)$. Indeed assume by contradiction that $\exists x \in \mbox{cl}_D(U)\setminus \mbox{cl}_X(U) $ . This means $\exists U^{(X)}_x$ such that $U^{(X)}_x\cap U = \emptyset$ and in particular $(U^{(X)}_x\cap D)\cap U= \emptyset$. If we set $U^{(D)}_x=(U^{(X)}_x\cap D)$, we obtain a contradiction because $(D\cap U^{(D)}_x)=\emptyset$.

Second we have that saying $\nexists U^{(X)}_x$ such that $U^{(X)}_x\subseteq \mbox{cl}_X(U)$ is equal to say that $\forall U^{(X)}_x $, $U^{(X)}_x\cap (\mbox{cl}_X(U))^c\neq\emptyset\ $ (the complementar of the closure in $X$ of $U$). Note that this last set is open, and so for density we have $\forall U^{(D)}_x $, $U^{(D)}_x\cap (\mbox{cl}_X(U))^c\neq\emptyset\ $.

I think this should work.

$\endgroup$
7
  • $\begingroup$ I'm fairly certain the statement is true. You definitely made an error on point 1. I think it is enough to show that $\mbox{cl}_D (U \cap D) = \mbox{cl}_X U \cap D$ for an open set $U$ and also that $\mbox{int}_D (A \cap D) = \mbox{int}_X A \cap D$ for a closed set A. $\endgroup$
    – ihaphleas
    Nov 15, 2013 at 22:39
  • $\begingroup$ @thatguy: No, (1) is correct. If $x\in\operatorname{int}_D\operatorname{cl}_D(U\cap D)$, then $x\in D$, so $x\notin U\cap D$ iff $x\notin U$. $\endgroup$ Nov 16, 2013 at 14:08
  • $\begingroup$ @BrianM.Scott: Hmm, seems you're right. $\endgroup$
    – ihaphleas
    Nov 16, 2013 at 15:02
  • $\begingroup$ @Brianm. Scott: You are right, the only thing that I do not completely demonstrate is the final statement. $\endgroup$ Nov 16, 2013 at 15:19
  • $\begingroup$ I have done an edit, so now the proof should be complete. (I hope). $\endgroup$ Nov 18, 2013 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .