Found a neat properties for dense and nowhere dense sets in functional analysis assignments recently. Suppose 
is a topological space, and 
is an arbitrary subset. We say that 
is dense if the closure is the whole space itself, i.e. 
; we say that is nowhere dense if the interior of its closure is empty, i.e. 
.
Now, we can also think about closures and interiors in terms of their complement. We shall denote by 
to be the complement of instead of the usual 
; the reason is obvious once we get to the main part, I promise.
Usually in topology, the interior and closure of a set , are defined to be the union of all open sets contained in , and the intersection of all closed sets containing respectively; in other words, they are the largest open set contained in , and the smallest closed set that contains respectively. In terms of complement, we have the interior 
and 
, this is not hard to prove.
If is dense, then the complement has empty interior. We have
This is not surprising and is pretty standard. On the other hand, we claim that is nowhere dense if and only if its complement contains an open dense subset of 
. This is slightly more work, we have
And since the interior itself is an open set, hence 
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!
Math Insanity 