Topology: An Inquiry-Based Approach

Let $X$ be a metric space and $A$ a subset of $X\text{.}$ To prove that $\bar{A}$ is a closed set, we will prove that $\bar{A}$ contains its limit points. Let $x\in {\bar{A}}^{\prime }\text{.}$ To show that $x\in \bar{A}\text{,}$ we proceed by contradiction and assume that $x\notin \bar{A}\text{.}$ This implies that $x\notin A$ and $x\notin A^{\prime }\text{.}$ Since $x\notin A^{\prime }\text{,}$ there exists a neighborhood $N$ of $x$ that contains no points of $A$ other than $x\text{.}$ But $A\subseteq \bar{A}$ and $x\notin \bar{A}\text{,}$ so it follows that $N\cap A=\mathrm{\varnothing }\text{.}$ This implies that there is an open ball $B\subseteq N$ centered at $x$ so that $B\cap A=\mathrm{\varnothing }\text{.}$ The fact that $x\in {\bar{A}}^{\prime }$ means that $B\cap \bar{A}$ contains a point $y$ in $\bar{A}$ different from $x\text{.}$ Since $B\cap A=\mathrm{\varnothing }\text{,}$ we must have $y\in A^{\prime }\text{.}$ But this, and the fact that $B$ is a neighborhood of $y\text{,}$ means that $B$ must contain a point of $A$ different than $y\text{.}$ But $B\cap A=\mathrm{\varnothing }\text{,}$ so we have reached a contradiction. We conclude that $x\in \bar{A}$ and ${\bar{A}}^{\prime }\subseteq \bar{A}\text{.}$ This shows that $\bar{A}$ is a closed set.

The proof that $\bar{A}$ is the smallest closed subset of $X$ that contains $A$ is left for the next activity.

In either case we have $x\in \bar{A}\cap \overline{X\setminus A}$ and so $\text{Bdry}(A)\subseteq \bar{A}\cap \overline{X\setminus A}\text{.}$

For the reverse implication, refer to the next activity.