Remark in "Lecture Notes on Elementary Topology and Differential Geometry" (Singer/Thorpe) In leading up to the Baire Category Theorem, the following definition was given in the aforementioned text:
"Let $S$ be a topological space. A subset $T$ of $S$ is "nowhere dense" if $ \bar T$ (The closure of $T$) contains no non-empty open set."
The following remark was given immediately after:
" $T \subset S$ is nowhere dense if and only if ($ \bar T$)' (the complement of $ \bar T$) is dense in $S$." 
(Note that the text's definition of a subset $A$ of $S$ being dense in $S$ is that $\bar A = S$)
My first attempt to prove this remark involved applying De Morgan's Laws to various unions and intersections. However, this became cumbersome very quickly. My second attempt involved a proof by contradiction, but this was not fruitful either. 
Any help in proving this remark would be appreciated. Thanks in advance. 
 A: Hint: Show that a subset $A$ is dense iff its intersection with every nonempty open set is nonempty. 
 Apply it with $A=S\setminus \bar T$. 
A: For any space there is a nice duality between interiors and closures:
$$\operatorname{int}(A) = (\overline{A^\complement})^\complement$$
and as $A$ nowhere dense means $\operatorname{int}(\overline{A})=\emptyset$
we conclude that for nowhere dense $A$:
$$\emptyset = (\overline{\overline{A}^\complement})^\complement$$
so that $$\overline{\overline{A}^\complement}= X$$ or otherwise put:
$\overline{A}^\complement$ is dense in $X$
Every statement can be reversed so we have equivalence of $A$ nowhere dense and the complement of its closure being dense.
A: Let $T$ be a nowhere dense set. Then if $U\subset S$ is open, $U\subset S\setminus T$. If $p$ is a limit point of $S$, we must have $p\in S\setminus T$, because if $V$ is a neighborhood of $p$, then $V$ does not intersect $T$. Therefore $\overline{S\setminus \overline T}=S$.
Suppose now that $S\setminus \overline T$ is dense in $S$, then $\overline{S\setminus\overline T}=S$. Suppose $U\subset \overline T$ is open. Then $U=T\cap A$ for some open $A$ in $S$. But $S=\overline{S\setminus{\overline T}}$, so $A\subset \overline{S\setminus{\overline T}}$. Since $\overline T\cap\overline{S\setminus{\overline T}}=\varnothing$, it follows that $U=\varnothing$.
