Activity 10.2.
Let \(X\) be a metric space.
(a)
Is \(\emptyset\) closed in \(X\text{?}\) Explain.
(b)
Is \(X\) closed in \(X\text{?}\) Explain.
Section Closed Sets in Metric Spaces
Recall that Definition 10.1 defines a closed set in a metric space to be a set whose complement is open. We have seen that both \(\emptyset\) and \(X\) are open subsets of \(X\text{.}\) We now ask the same question, this time in terms of closed sets.
Note that a subset of a metric space can be both open and closed. We call such sets clopen (for closed-open). When we discussed open sets, we saw that an arbitrary union of open sets is open, but that an arbitrary intersection of open sets may not be open (although a finite intersection of open sets is open). Since closed sets are complements of open sets, we should expect a similar result for closed sets.
Activity 10.3.
Let \(X = \R\) with the Euclidean metric. Let \(A_n = \left[\frac{1}{n}, 1-\frac{1}{n}\right]\) for each \(n \in \Z^+\text{,}\) \(n \geq 2\text{.}\)
(a)
What is \(\bigcup_{n \geq 2} A_n\text{?}\) A proof is not necessary.
(b)
Is \(\bigcup_{n \geq 2} A_n\) closed in \(\R\text{?}\) Explain.
Activity 10.3 shows that an arbitrary union of closed sets is not necessarily closed. However, the following theorem tells us what we can say about unions and intersections of closed sets. The results should not be surprising given the relationship between open and closed sets.
Theorem 10.4.
Let \(X\) be a metric space.
- Any intersection of closed sets in \(X\) is a closed set in \(X\text{.}\)
- Any finite union of closed sets in \(X\) is a closed set in \(X\text{.}\)
Proof.
Let \(X\) be a metric space. To prove part 1, assume that \(\{C_{\alpha}\}\) is a collection of closed sets in \(X\) for \(\alpha\) in some indexing set \(I\text{.}\) DeMoivre’s Theorem shows that
\begin{equation*}
X \setminus \bigcap_{\alpha \in I} C_{\alpha} = \bigcup_{\alpha \in I} X \setminus C_{\alpha}\text{.}
\end{equation*}
The latter is an arbitrary union of open sets and so it an open set. By definition, then, \(\bigcap_{\alpha \in I} C_{\alpha}\) is a closed set.
For part 2, assume that \(C_1\text{,}\) \(C_2\text{,}\) \(\ldots\text{,}\) \(C_n\) are closed sets in \(X\) for some \(n \in \Z^+\text{.}\) To show that \(C = \bigcup_{k=1}^n C_k\) is a closed set, we will show that \(X \setminus C\) is an open set. Now
\begin{equation*}
X \setminus \bigcup_{i=1}^n C_{i} = \bigcap_{i=1}^n X \setminus C_{i}
\end{equation*}
is a finite intersection of open sets, and so is an open set. Therefore, \(\bigcup_{i=1}^n C_{i}\) is a closed set.
