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 $\emptyset$.
Best Answer
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 $V\subseteq \overline N^c\subseteq N^c$.