Topology: An Inquiry-Based Approach

Let $X$ be a topological 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 set $O\subseteq N$ centered at $x$ so that $O\cap A=\mathrm{\varnothing }\text{.}$ The fact that $x\in {\bar{A}}^{\prime }$ means that $O\cap \bar{A}$ contains a point $y$ in $\bar{A}$ different from $x\text{.}$ Since $O\cap A=\mathrm{\varnothing }\text{,}$ we must have $y\in A^{\prime }\text{.}$ But the fact that $O$ is a neighborhood of $y$ means that $O$ must contain a point of $A$ different than $y\text{,}$ which contradicts the fact that $O\cap A=\mathrm{\varnothing }\text{.}$ 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.