Showing that the set of interior points of a set in a metric space is the largest open set.

Question

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.

Answer 1

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 $\text{int}(M)$.

Answer 2

This isn't really difficult, if you write down the definition of the interior. Let $(X,d)$ be a metric space and $M\subset X$. Then

\begin{align*} \text{Int}(M) = \bigcup \{ U \,| U \text{ is open }, U \subset M\} \end{align*}

This means automatically, that Int($M$) is open, since arbitrary unions of open sets are open (why?). Also clearly Int($M$)$\subset M$, since it is the union of sets, which are all subsets of $M$.

Now, saying that Int($M$) is the biggest open set contained in $M$, is just the same as saying, whenever $U\subset M$ and $U$ open, then $U \subset \text{Int}(M)$. But if you look at the definition of Int($M$), you see that every such $U$ was part of the union defining Int($M$).

Answer 3

Fix a set $M$. Let $O$ be any open set such that $O \subseteq M$. I claim that $O \subseteq \operatorname{int}(M)$. To see this, let $x \in O$. As $O$ is open, $x$ is an interior point of $O$ so there is some $r>0$ such that $B(x,r) \subseteq O$. But as $O \subseteq M$ we also have that $B(x,r) \subseteq M$, which means that $x \in \operatorname{int}(M)$. This shows the inclusion.

The only thing we need to say that $\operatorname{int}(M)$ is the maximal open subset of $M$ is that $\operatorname{int}(M)$ is itself open, and this follows by the standard argument that all sets of the form $B(x,r)$ are open.