Interior of closure of open set in topological space

Question

If $U$ is an open set in a topological space, is it true that $U$ is the interior of the closure of itself?

If $U$ is open, it must be the interior of itself. But is it the interior of the closure of itself? In the closure, we include all points $x$ such that any open set containing $x$ also contains a point in $U$. In the interior of the closure, we take the union of all open sets contained in the closure.

If that were always so, the term "regular open set" would not have been invented. — Daniel Fischer, Sep 15 '13 at 19:44

score 2 · Accepted Answer · answered Sep 15 '13 at 19:44

2

Hint: What is the interior of the closure of $\mathbb{R}\setminus\{0\}$?

answered Sep 15 '13 at 19:44

Michael Albanese

99,526

score 2 · Answer 2 · answered Sep 15 '13 at 19:45

2

Not quite. Consider a punctured disc. That is consider $U=\{x\in\mathbb{R}^{2}|0<\|x\|<1\}$.

answered Sep 15 '13 at 19:45

user71352

13,038

Interior of closure of open set in topological space

2 Answers2

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