Let $(X, d)$ be a metric space, and let $M \subset X.$ I'm trying to show that the set of interior points of $M$ ($\text{int}(M)$) is the largest open set contained in $M$.
My way of thinking I should approach this is by first proving $\text{int}(M) \subset M$, then showing $\text{int}(M)$ is open, and then for any other open set $A \subset M$, $A \subset \text{int}(M)$. I'm a little lost on where to start so some initial direction might be useful.
Best Answer
Suppose $x\in A$. Since $A$ is open, there is an open ball $B_{\epsilon}(x)\subset A$. But, $A$ is contained in $M$, so that $B_{\epsilon}(x)\subset M$. Thus, $x\in \text{int}(M)$, since $x\in M$ and we found an open ball surrounding $x$ that lies entirely in $M$. It follows that $A\subset \text{int}(M)$.
The fact that $\text{int}(M)\subset M$ is clear, since by definition, $\text{int}(M)=\{x\in X:\exists\epsilon>0\text{ such that }x\in B_{\epsilon}(x)\subset M\}$.
To see that $\text{int}(M)$ is open, suppose $x\in \text{int}(M)$. Then, there is an open ball $x\in B_{\epsilon}(x)\subset M$. The claim is that $B_{\epsilon}(x)$ is in fact contained in $\text{int}(M)$. To see this, fix $z\in B_{\epsilon}(x)$. Let $r=d(x,z)$. Then check that, $B_{r}(z)\subset B_{\epsilon}(x)\subset M$. But this then means that $z\in\text{int}(M)$. Hence, $B_{\epsilon}(x)\subset \text{int}(M)$. This proves that $\text{int}(M)$ is open, since for an arbitrary $x\in \text{int}(M)$, you found a ball containing $x$ that lies entirely within $\text{int}(M)$.