Let $X$ be a topological space, let $Y$ be an open subspace of $X$. Suppose that $S \subseteq Y$ is nowhere dense in $Y$, that is, $\operatorname{int}_Y(\operatorname{cl}_Y(S)) = \emptyset$. How to prove that $S$ is nowhere dense in $X$?

I have not made much progress so far. As a starting point, the assumption $\operatorname{int}_Y(\operatorname{cl}_Y(S)) = \emptyset$ can be written in terms of closures and interiors in $X$, but this does not seem to be of much help so far..

  • $\begingroup$ $\operatorname{int}(A\cap B) = \operatorname{int}(A) \cap \operatorname{int}(B)$. Let $A = \overline{S}$, and $B = Y$. So $\operatorname{int}(\overline{S})$ is entirely contained in $X\setminus Y$, hence? $\endgroup$ – Daniel Fischer Oct 22 '13 at 16:02
  • $\begingroup$ I'm not sure what you mean. I suppose your idea is to prove then that $\operatorname{int}(\bar{S})$ is contained in $Y$, but I don't see how to conclude that. $\endgroup$ – spin Oct 22 '13 at 16:27
  • $\begingroup$ The idea is that then $\operatorname{int}(\overline{S}) \cap \overline{Y} = \varnothing$. But $\overline{S}\subset\overline{Y}$. $\endgroup$ – Daniel Fischer Oct 22 '13 at 17:23

I recommend an alternate characterization of nowhere dense. The following is a good exercise to prove.

Lemma: Given a subset $S$ of a topological space $Z,$ the following are equivalent:

  • $S$ is nowhere dense in $Z$.
  • For every non-empty open $V\subseteq Z$, there is a non-empty open $W\subseteq V$ such that $S\cap W=\emptyset.$

That makes the result almost trivial. Take any non-empty open subset $U$ of $X$. If $U\cap Y=\emptyset,$ then $U\cap S=\emptyset,$ so suppose not. Then $V=U\cap Y$ is a non-empty open subset of $Y$, and so, bearing in mind that $S$ is nowhere dense in $Y$ and that $Y$-open sets are $X$-open, what can we conclude?

  • $\begingroup$ There is something weird about the lemma, $W$ does not depend on $V$? Do you mean $W \subseteq V$? $\endgroup$ – spin Oct 22 '13 at 16:29
  • $\begingroup$ Oops! Yes, that was supposed to be $W\subseteq V$. Fixed. $\endgroup$ – Cameron Buie Oct 22 '13 at 16:30
  • $\begingroup$ Thanks, this works. With this idea it's very easy to prove. $\endgroup$ – spin Oct 22 '13 at 16:39
  • 1
    $\begingroup$ Excellent! Glad I could help (and that you caught my error). $\endgroup$ – Cameron Buie Oct 22 '13 at 16:42

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