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I am wondering if this is generally true for any topology. I think there might be counter examples, but I am having trouble generating them.

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The set $(0,1) \cup (1,2)$ is a counterexample.

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No. A quick way to verify counterexamples is the following observation: if $U$ is the interior of a closed set $C$, then $U$ is also the interior of $\overline{U}$. Indeed, since $C$ is closed and $U\subseteq C$, $\overline{U}\subseteq C$, so the interior of $\overline{U}$ is contained in the interior of $C$. But $U$ is contained in the interior of $\overline{U}$ (since it is open and contained in $\overline{U}$) and equal to the interior of $C$, so $U$ must be equal to the interior of $\overline{U}$.

We can now give an example: take $U=(0,1)\cup(1,2)$ as a subset of $\mathbb{R}$. Since $\overline{U}=[0,2]$ has interior $(0,2)\neq U$, $U$ cannot be the interior of any closed set. In general, a set which is the interior of a closed set (equivalently, of its closure) is called a regular open set.

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Since the complement of an open set is closed (and vice versa), and since the complement of the interior is the closure of the complement, we can rephrase your question equivalently as:

Is every closed set the closure of some open set?

This immediately suggests a counterexample: any singleton (i.e. a set containing only one point) is closed in $\mathbb R^n$ (with the usual Euclidean topology), but has no non-empty open subsets that it could be the closure of.

Conversely, the complement of any singleton (i.e. $\mathbb R^n \setminus \{x\}$ for any $x \in \mathbb R^n$) provides a counterexample to your original claim, being an open set that cannot be the interior of any closed set.

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In the Sierpinski topology $\{X,\emptyset,\{0\}\}$, the set $\{0\}$ is an open set that isn't the interior of any closed set.

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  • $\begingroup$ Because the only closed sets are $X, \emptyset$ and $\{1\}$, which have interiors $X, \emptyset, \emptyset$ respectively. $\endgroup$ – Henno Brandsma Feb 5 '17 at 6:11
  • $\begingroup$ Downvote now after 11 positive votes shows that mathematics changes with time ;). $\endgroup$ – Nosrati Dec 10 '17 at 20:39
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A punctured disk and a slit disk are easy examples in the plane.

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Every open, dense set is a counterexample. In $\mathbb{R}$ you can use $$\bigcup_{n \in \mathbb{N}} (q_n - 1/2^n, q_n + 1/2^n),$$ where $\mathbb{Q} = \{q_0, q_1, \ldots\}$ are the rationals.

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