Topology: An Inquiry-Based Approach

Let $(X,d)$ be a metric space, and let $A$ be a subset of $X\text{.}$ We need to prove that $\text{Int}(A)$ is an open set in $X\text{,}$ and that $\text{Int}(A)$ is the largest open subset of $X$ contained in $A\text{.}$ First we demonstrate that $\text{Int}(A)$ is an open set. Let $a\in \text{Int}(A)\text{.}$ Then $a$ is an interior point of $A\text{,}$ so $A$ is a neighborhood of $a\text{.}$ This implies that there exists an $ϵ>0$ so that $B(a,ϵ)\subseteq A\text{.}$ But $B(a,ϵ)$ is a neighborhood of each of its points, so every point in $B(a,ϵ)$ is an interior point of $A\text{.}$ It follows that $B(a,ϵ)\subseteq \text{Int}(A)\text{.}$ Thus, $\text{Int}(A)$ is a neighborhood of each of its points and, consequently, $\text{Int}(A)$ is an open set.

The proof that $\text{Int}(A)$ is the largest open subset of $X$ contained in $A$ is left for the next activity.