# Existence of a dense proper subset with non-empty interior?

1. What is an example of a topological (or particularly a metric) space $X$ having a proper non-open dense subset with non-empty interior?
2. Is it possible for a closed dense subset of a topological space to have an empty interior?
3. Finally: on a similar note: I've noticed that $\mathbb{Q}$ doesn't work, why?

(I will post my solution below as a Q& A type problem/solution:)

Example of a closed dense subset with non-empty interior

As a counter example of a proper dense subset of a set of topological space with non-empty interior consider $$D:=(-\infty,0]\cup (\mathbb{Q}\cap(0,1)) \cup[1,\infty)$$ in $\mathbb{R}$ with its usual topology, then since:

• $(-\infty,0]$ and $\cup[1,\infty)$ are closed in $\mathbb{R}$, they must equal to their closure.
• $\mathbb{Q}\cap(0,1)$ is not open and dense in $[0,1]$ since: $\mathbb{Q}$ is not open and is density $\mathbb{R}$, taking intersections with $(0,1)$ yields the conclusion.

therefore we can conclude that $D$ is not closed, dense in $\mathbb{R}$ and a proper subset thereof.

Moreover, the subset $(\infty,0)\cup(0,\infty)$ is open in $D$ (and infact is the largest such subset, therefore is equal to its interior), hence the interior of $D$ is non-empty.

Concerning the existence of a closed dense subset with empty interior:

By the definition of density the closure of $D$ must be equal to $X$. However the axioms of a topological space assure that $X$ is closed in $X$ and by the since $D$ was assuemd to be closed also it must be equal to its closure, therefore $X=D$ if $D$ is closed and dense in $X$.

There is only one possibility of a topological space $X$ which has a closed dense subset $D$ with empty interior, that is $X$ itself must be the empty set. At this point it becomes a question of convention! Most people require that the initial object in the category of topological spaces be the one-point space, therefore the empty set (though it does satisfy the axioms of a topology using the empty-topology thereon) is not generally regarded as begin a topological space. Hence under this convention there is no such space (if you reject this very standard convention (and almost definition at this point) then our non-mainstream example would be the only one which is the the empty topological space with the topology $\{ \emptyset \}$). (on a personal note, I require topological space to be non-empty).

Concerning the empty interior of $\mathbb{Q}$:

The empty interior follows from $\mathbb{R}-\mathbb{Q}$ is dense in $\mathbb{R}$ and remarked is not from $\mathbb{Q}$'s density in $\mathbb{R}$.

The reason being that in every open ball around a rational number there exists and irrational number, hence there cannot be an open ball in $\mathbb{R}$ properly contained in $\mathbb{Q}$. Since the open balls generate the metric topology on $\mathbb{R}$, then there cannot be a (non-empty) open subset of $\mathbb{Q}$ contained within $\mathbb{Q}$. Now using the definition of interior as the union of all open subsets of a set, this implies the interior of $\mathbb{Q}$ is empty.

• Sorry that was a typo I meant $\{ \emptyset \}$, thanks for pointing that out :) – AIM_BLB Jun 2 '15 at 18:17