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 \begin{equation} D:=(-\infty,0]\cup (\mathbb{Q}\cap(0,1)) \cup[1,\infty) \end{equation} 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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.