Definition 13.7.
The closure of a subset \(A\) of a topological space \(X\) is the set
\begin{equation*}
\overline{A} = A \cup A'\text{.}
\end{equation*}
Section Closure in Topological Spaces
Once we have a definition of limit point, we can define the closure of a set just as we did in metric spaces.
In other words, the closure of a set is the collection of the elements of the set and the limit points of the set. The following theorem is the analog of the theorem in metric spaces about closures.
Theorem 13.8.
Let \(X\) be a topological space and \(A\) a subset of \(X\text{.}\) The closure of a \(A\) is a closed set. Moreover, the closure of \(A\) is the smallest closed subset of \(X\) that contains \(A\text{.}\)
Proof.
Let \(X\) be a topological space and \(A\) a subset of \(X\text{.}\) To prove that \(\overline{A}\) is a closed set, we will prove that \(\overline{A}\) contains its limit points. Let \(x \in \overline{A}'\text{.}\) To show that \(x \in \overline{A}\text{,}\) we proceed by contradiction and assume that \(x \notin \overline{A}\text{.}\) This implies that \(x \notin A\) and \(x \notin A'\text{.}\) Since \(x \notin A'\text{,}\) there exists a neighborhood \(N\) of \(x\) that contains no points of \(A\) other than \(x\text{.}\) But \(A \subseteq \overline{A}\) and \(x \notin \overline{A}\text{,}\) so it follows that \(N \cap A = \emptyset\text{.}\) This implies that there is an open set \(O \subseteq N\) centered at \(x\) so that \(O \cap A = \emptyset\text{.}\) The fact that \(x \in \overline{A}'\) means that \(O \cap \overline{A}\) contains a point \(y\) in \(\overline{A}\) different from \(x\text{.}\) Since \(O \cap A = \emptyset\text{,}\) we must have \(y \in A'\text{.}\) 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 \(O \cap A = \emptyset\text{.}\) We conclude that \(x \in \overline{A}\) and \(\overline{A}' \subseteq \overline{A}\text{.}\) This shows that \(\overline{A}\) is a closed set.
The proof that \(\overline{A}\) is the smallest closed subset of \(X\) that contains \(A\) is left for the next activity.
Activity 13.5.
Let \((X,d)\) be a topological space, and let \(A\) be a subset of \(X\text{.}\)
(a)
What will we have to show to prove that \(\overline{A}\) is the smallest closed subset of \(X\) that contains \(A\text{?}\)
(b)
Suppose that \(C\) is a closed subset of \(X\) that contains \(A\text{.}\) To show that \(\overline{A} \subseteq C\text{,}\) why is it enough to demonstrate that \(A' \subseteq C\text{?}\)
(c)
If \(x \in A'\text{,}\) what can we say about \(x\text{?}\)
(d)
Complete the proof that \(\overline{A} \subseteq C\text{.}\)
One consequence of Theorem 13.8 is the following.
Corollary 13.9.
A subset \(C\) of a topological space \(X\) is closed if and only if \(C = \overline{C}\text{.}\)
