Theorem 10.8.
Let \(C\) be a subset of a metric space \(X\text{,}\) and let \(C'\) be the set of limit points of \(C\text{.}\) Then \(C\) is closed if and only if \(C' \subseteq C\text{.}\)
Section Limit Points and Closed Sets
There is a connection between limit points and closed sets. The open set \((1,2)\) in \((\R, d_E)\) does not contain all of its limit points or any of its boundary points, while the closed set \([1,2]\) contains all of its boundary and limit points. This is an important attribute of closed sets. Recall that for a limit point \(x\) of a subset \(A\) of a metric space \(X\text{,}\) every neighborhood of \(x\) contains a point in \(A\) different from \(x\text{.}\) We can make the neighborhoods as small as we like so, in a sense, the limit points of \(A\) that are not in \(A\) are the points in \(X\) that are arbitrarily close to the set \(A\text{.}\) We denote the set of limit points of \(A\) as \(A'\text{,}\) and the limit points of a set can tell us if the set is closed.
Proof.
Let \(X\) be a metric space, and let \(C\) be a subset of \(X\text{.}\) First we assume that \(C\) is closed and show that \(C\) contains all of its limit points. Let \(x \in X\) be a limit point of \(C\text{.}\) We proceed by contradiction and assume that \(x \notin C\text{.}\) Then \(x \in X \setminus C\text{,}\) which is an open set. This implies that there is an \(\epsilon \gt 0\) so that \(B(x, \epsilon) \subseteq X \setminus C\text{.}\) But then this neighborhood \(B(x, \epsilon)\) contains no points in \(C\text{,}\) which contradicts the fact that \(x\) is a limit point of \(C\text{.}\) We conclude that \(x \in C\) and \(C\) contains all of its limit points.
The converse of the result we just proved is the subject of the next activity.
Activity 10.8.
Let \(C\) be a subset of a metric space \(X\text{,}\) and let \(C'\) be the set of limit points of \(C\text{.}\) In this activity we prove that \(C\) is closed if \(C\) contains all of its limit points. So assume \(C' \subseteq C\text{.}\)
(a)
What do we need to do to show that \(C\) is closed?
(b)
If we proceed by contradiction to prove that \(C\) is closed, we assume that \(C\) is not closed. What does this tell us about \(X \setminus C\text{?}\)
(c)
What does the conclusion of part (b) tells us?
(d)
How does the result of (c) contradict the assumption that \(C\) contains all of its limit points?
