# Dense and Nowhere Dense Sets

Found a neat properties for dense and nowhere dense sets in functional analysis assignments recently. Suppose $(X,\tau)$ is a topological space, and $A \subseteq X$ is an arbitrary subset. We say that $A$ is dense if the closure is the whole space itself, i.e. $\bar{A} = X$; we say that $A$ is nowhere dense if the interior of its closure is empty, i.e. $\mathring{\bar{A}} = \emptyset$.

Now, we can also think about closures and interiors in terms of their complement. We shall denote by $X \setminus A$ to be the complement of $A$ instead of the usual $A^C$; the reason is obvious once we get to the main part, I promise.

Usually in topology, the interior and closure of a set $A$, are defined to be the union of all open sets contained in $A$, and the intersection of all closed sets containing $A$ respectively; in other words, they are the largest open set contained in $A$, and the smallest closed set that contains $A$ respectively. In terms of complement, we have the interior $\mathring{A} = X \setminus \bar{A}$ and $\bar{A} = (X \setminus A)^{\circ}$, this is not hard to prove.

If $A$ is dense, then the complement has empty interior. We have

$\bar{A} = X \iff X \setminus \bar{A} = \emptyset \iff (X \setminus A)^{\circ} = \emptyset$

This is not surprising and is pretty standard. On the other hand, we claim that $A$ is nowhere dense if and only if its complement contains an open dense subset of $X$. This is slightly more work, we have

$\mathring{\bar{A}} = \emptyset \iff \bar{X \setminus \bar{A}} = X \iff \bar{(X \setminus A)^{\circ}} = X \iff (X \setminus A)^{\circ} = X$

And since the interior itself is an open set, hence $(X \setminus A)^{\circ}$ is the open dense subset we desired.

This is pretty useful result when trying to prove Baire Category’s Theorem, or for an application of the Baire’s Category Theorem. All the best doing your assignments!