# How does definition of nowhere dense imply not dense in any subset?

In some topological space $X$, a set $N$ is nowhere dense iff $\text{Int}\left(\overline{N}\right)=\emptyset$, where Int is the interior, and overbar is closure.

How can I show this is equivalent to the statement "any non-empty open subset of $X$ contains an open non-empty subset containing no elements of $N$."?

I can use the following equivalences: $N\text{ is nowhere dense}\iff\overline{N}\text{ is nowhere dense}\iff\text{Ext}\left(N\right)\text{ is dense in }X\;\iff\overline{N}^{c}\text{ is dense in }X$

My attempt:

1) I assumed the quoted statement is false - that there is a non-empty open set $V$ in which $N$ is dense. It follows that $V\subset\overline{N}$ . On the other hand, N is nowhere dense, so $V\subset\overline{\text{Ext}\left(N\right)}$ also holds. I concluded that $V\subset\partial N$ , but I don't see what this achives..

2) I think I might have a proof - is the following correct?

$\text{Int}\left(N\right)=\emptyset\iff\text{Int}\left(\overline{N}\right)=\emptyset$, so we can in particular take $N_{1}=\overline{N}\setminus\partial N$, and have $\text{Int}\left(N_{1}\right)=\emptyset$. But $\text{Int}\left(N_{1}\right)=N_{1}$, so $N_{1}=\emptyset$. Additionally, $\text{Int}\left(N_{1}\right)$ is the greatest subset of $N_{1}$ (itself), so any subset of it must be empty. The boundary contains no open subsets, so the family of nowhere dense sets whose closure is $\overline{N}$ are just closed sets: they're all unions of the empty set with a subset of a the boundary - a closed set. This means the greatest open subset of nowhere dense set is $\emptyset$.

• Do you mean "any non-empty open subset of $X$ contains an open non-empty subset containing no elements of $N$." ? – Stefan Hamcke Aug 20 '13 at 15:14
• Yes, thank you! Corrected. – Sai Aug 20 '13 at 15:18

Note that $\overline N^c \text{ is dense in }X \Longleftrightarrow \overline N^c\cap U\neq \emptyset \text{ for all }U\neq \emptyset \text{ open in }X$. Note further that $\overline N^c$ is open. Now, let $U$ be a non-empty open subset of $X$, by the above $V=\overline N^c\cap U$ is a non-empty open subset of $U$ containing no elements of $N$, because $V\subseteq \overline N^c\subseteq N^c$.

• Chose this because it's constructive – Sai Aug 20 '13 at 15:43
• @Sai: You can actually "upvote" any of the answers that you like. You can only accept one answer, though. (I note you haven't done that, yet. Did you mean to?) – Cameron Buie Aug 20 '13 at 16:44
• @Cameron I did upvote them all. Forgot to accept though, thanks! – Sai Aug 20 '13 at 17:18

As written, you're not quite there. It should be "any non-empty open subset of $X$ contains a non-empty open subset containing no elements of $N$."

On the one hand, then, suppose $\operatorname{Ext}(N)$ is dense in $X$, and take any non-empty open $U$. By definition of "dense," there is some $x\in\operatorname{Ext}(N)$ such that $x\in U$. What does this allow us to conclude about $\operatorname{Int}(N),$ since $U$ was an arbitrary open set?

On the other hand, suppose that $\operatorname{Ext}(N)$ is not dense in $X,$ so that there is some non-empty open $U$ disjoint from $\operatorname{Ext}(N).$ Then $U^c$ is a closed set containing $\operatorname{Ext}(N),$ and so $\overline{\operatorname{Ext}(N)}\subseteq U^c,$ meaning that $U\subseteq\overline{\operatorname{Ext}(N)}^c=\operatorname{Int}(N).$

If you know that $V\subseteq \partial N$, it follows that $V\subseteq\overline N$ which is impossible since $\text{int}\overline N$ is empty.

Edit: There is no need to aim at a contradiction, a direct way is more appropiate here: You assumed that int$\bar N$ is empty (by definition of nowhere dense). Then you assumed that $N\cap V$ is dense in $V$, which implies $V\subseteq\bar N$ since the closure of $N\cap V$ relative to $V$ is the same as $V\cap(\overline{ N\cap V})$. (Whenever $V$ is open and $N$ is any set, then the closure of $N\cap V$ relative to $V$ is even $V\cap\bar N.$) Since $V$ is open and a subset of $\bar N$, it follows that $V$ is a subset of int$\bar N$, thus empty. It follows that there is no non-empty open $V$ where the intersection $N\cap V$ is dense.

• Isn't $\text{Int}\left(N\right)=\emptyset=\overline{N}\setminus\partial N$? Maybe i'm being dense here, but where is the contradiction? After all $V\subset\partial N\subset\overline{N}$ so $V\nsubseteq\text{Int}\left(N\right)$ and $\text{Int}\left(N\right)$ can stay empty – Sai Aug 20 '13 at 15:54
• @Sai: You assumed that int$\bar N$ is empty (by definition of nowhere dense). Then you assumed that $N\cap V$ is dense in $V$, which you correctly identified to be equal to $V\subseteq\bar N$ since the closure of $N\cap V$ relative to $V$ is the same as $\bar N\cap V$. (This is true whenever $V$ is open and $N$ is any set, but need not be true for non-open $V$.) Since $V$ is open and a subset of $\bar N$, it follows that $V$ is a subset of int$\bar N$. – Stefan Hamcke Aug 20 '13 at 16:10
• Thank you, I'm a little less dense now :) – Sai Aug 20 '13 at 16:14