Prove that $Y$ is the closure of some open set if and only if $Y$ is the closure of its interior.

Question

Prove that $Y$ is the closure of some open set if and only if $Y$ is the closure of its interior.

$\implies: $ Let $Y=\overline Z$ for some open set $Z$. Then $Y$ is the smallest closed set containing $Z$. Since $\mathring Y$ is the largest open set contained in $Y$, therefore $Z\subseteq \mathring Y$. However, since $\overline{\mathring Y}$ is the smallest closed set containing $\mathring Y$, we have that $\overline{\mathring Y} \subseteq Y$, but being that $Y$ is the smallest closed set containing $Z$ it is also the case that $Y\subseteq \overline{\mathring Y}$ and hence $Y= \overline{\mathring Y}$.

$\Longleftarrow:$ Assume that $Y=\overline{\mathring Y}$. Since $\mathring Y$ is the largest open set contained in $Y$, for any other open set $Z$ in $Y$ we must have that $Z\subseteq \mathring Y$. And since $\overline Z$ is the smallest closed set containing $Z$, we have $\overline Z \subseteq \overline{\mathring Y}$. However, since $\overline{\mathring Y}$ is the smallest closed set containing $\mathring Y$, it is also true that $\overline{\mathring Y} \subseteq \overline Z$, so $\overline Z = \overline{\mathring Y}=Y$. $\Box$

Is this correct, and can this be simplified at all? The two directions seem largely similar in reasoning.

This problem confused me for a while since it was basically equivalent to saying, "Y is the smallest closed set containing an open set Z if and only if Y is the smallest closed set containing the largest open set contained in Y" which was hard for me to parse at first.

Answer 1

The proof of the forward direction is fine. After showing $Z\subseteq \mathring Y$ in the first paragraph, you can alternatively conclude by writing $$Y = \overline{Z} \subseteq \overline{\mathring Y} \subseteq \overline{Y} = Y$$ which proves $\overline{\mathring Y} = Y$. In the second paragraph, it would have been enough to note that since $\mathring Y$ is open, the reverse direction is proved.