Can a Closed Set Have No Nonempty Open Subsets?

Yes. In the real line, the Cantor set is closed and contains no nonempty open interval. Equivalently, its interior is empty. The singleton $\{0\}$ is a simpler example, but the Cantor set shows that a closed set can be much more complicated and still have no interior.

The phrase “open subset” needs one qualification: open relative to which space? The usual interpretation is open in the ambient space, such as $\mathbb{R}$. Under that interpretation, the examples below answer the question.

What “no open subsets” means

For a subset $F$ of a topological space $X$, its interior, written $\operatorname{int}(F)$, is the union of all open subsets of $X$ contained in $F$. Thus

$$\operatorname{int}(F)=\varnothing$$

means that $F$ contains no nonempty open subset of $X. The empty set is open in every topology, so literally no open subsets is impossible for any set.

A closed set with empty interior is often called nowhere dense when its closure also has empty interior. Since a closed set is equal to its closure, every closed set with empty interior is nowhere dense.

A simple example: a singleton

Take $F=\{0\}\subseteq\mathbb{R}$. It is closed: its complement $(-\infty,0)\cup(0,\infty)$ is open. But no nonempty open interval fits inside $\{0\}$, so

$$\operatorname{int}(\{0\})=\varnothing.$$

More generally, every finite subset of $\mathbb{R}$ is closed and has empty interior. A finite set is closed because its complement is open, and any nonempty open interval contains infinitely many points.

A less trivial example: the Cantor set

Start with the interval $C_0=[0,1]$. Remove its open middle third, $(1/3,2/3)$, leaving two closed intervals. From each remaining interval, remove its open middle third; repeat this process indefinitely. The Cantor set $C$ is what remains after all stages:

$$C=\bigcap_{n=0}^{\infty} C_n,$$

where $C_n$ is the union of the closed intervals left after stage $n$.

Why it is closed

Each $C_n$ is a finite union of closed intervals, hence is closed in $\mathbb{R}$. An arbitrary intersection of closed sets is closed, so $C$ is closed.

Why its interior is empty

At stage $n$, the remaining intervals each have length $3^{-n}$. Suppose, for a contradiction, that $C$ contained a nonempty open interval $(a,b)$. Choose $n$ large enough that $3^{-n}<b-a$. Since $(a,b)\subseteq C\subseteq C_n$, the interval $(a,b)$ would have to lie inside one of the component intervals of $C_n$: distinct component intervals are separated by gaps. But every component interval has length $3^{-n}$, which is smaller than $b-a$. That is impossible.

Therefore no nonempty open interval is contained in $C$, and $\operatorname{int}(C)=\varnothing$.

For example, $C$ contains $0$ and $1$, as well as many other points, so it is not just a finite or isolated-point example. Yet it still contains no interval.

The relative-topology distinction

If “open subset of $F$” means open in the subspace topology on $F$, then every subset is being judged using a different notion of openness. In particular, $F$ itself is open as a subset of $F$, because $F=F\cap\mathbb{R}$ and $\mathbb{R}$ is open in itself. So a nonempty set cannot have no nonempty open subsets relative to itself.

For the Cantor set, the statement is: it has no nonempty subset that is open in the ambient real line and contained in the Cantor set. It does have nonempty relatively open subsets—for instance, $C$ itself.

Common mistakes

• Using the rationals as an example: $\mathbb{Q}$ is dense in $\mathbb{R}$ but is not closed. Its closure is all of $\mathbb{R}$.
• Confusing “closed” with “having interior”: closed sets need not contain any open interval. A singleton and the Cantor set are both closed with empty interior.
• Forgetting the ambient space: a set can have empty interior in $\mathbb{R}$ while having nonempty open subsets in its own subspace topology.
• Taking “no open subsets” literally: the empty set is always open; the useful condition is “no nonempty open subsets,” or equivalently “empty interior.”