Let $E$ be a subset of $X$. Then $E$ is open if and only if $E = \text{int}(E)$.

Question

Let $E$ be a subset of $X$. Then $E$ is open if and only if $E = \text{int}(E)$. In other words, $E$ is open if and only if for every $x\in E$, there exists an $r > 0$ such that $B(x,r)\subseteq E$.

MY ATTEMPT

It is known that $E$ is open iff $\partial E\cap E = \varnothing$. Given that $X = \text{int}(E)\cup\partial E\cup\text{ext}(E)$ and $E\cap\text{ext}(E) = \varnothing$, we conclude that \begin{align*} E = E\cap X & = E\cap(\text{int}(E)\cup\partial E\cup\text{ext}(E))\\\\ & = (E\cap\text{int}(E))\cup(E\cap\partial E)\cup(E\cap\text{ext}(E))\\\\ & = E\cap\text{int}(E)\cup\varnothing\cup\varnothing = E\cap\text{int}(E) \end{align*} whence we conclude that $E\subseteq\text{int}(E)$. On the other hand, one has that $\text{int}(E)\subseteq E$.

Indeed, if $x\in\text{int}(E)$, there exists $r > 0$ such that $x\in B(x,r)\subseteq E$.

Hence we conclude that $E = \text{int}(E)$, and we are done.

Can someone please verify if I am reasoning correctly? Any other approach is welcome.

Answer 1

Your proof is correct, given that you have established the identities that you listed at the beginning of your proof.

Well done!