Section Unions and Intersections of Open Sets
Once we have defined open sets we might wonder about what happens if we take a union or intersection of open sets.
Activity 8.4.
(a)
Let \(A = (-2,1)\) and \(B = (-1,2)\) in \((\R, d_E)\text{.}\)
(i)
Is \(A \cup B\) open? Explain.
(ii)
Is \(A \cap B\) open? Explain.
(b)
Let \(X = \R\) with the Euclidean metric. Let \(A_n = \left(1-\frac{1}{n}, 1+\frac{1}{n}\right)\) for each \(n \in \Z^+\text{.}\)
(i)
What is \(\bigcup_{n \geq 1} A_n\text{?}\) A proof is not necessary.
(ii)
Is \(\bigcup_{n \geq 1} A_n\) open in \(\R\text{?}\) Explain.
(iii)
What is \(\bigcap_{n \geq 1} A_n\text{?}\) A proof is not necessary.
(iv)
Is \(\bigcap_{n \geq 1} A_n\) open in \(\R\text{?}\) Explain.
Activity 8.4 demonstrates that an arbitrary intersection of open sets is not necessarily open. However, there are some things we can say about unions and intersections of open sets.
Theorem 8.4.
Let \(X\) be a metric space.
Any union of open sets in \(X\) is an open set in \(X\text{.}\)
Any finite intersection of open sets in \(X\) is an open set in \(X\text{.}\)
Proof.
Let
\(X\) be a metric space. To prove part 1, assume that
\(\{O_{\alpha}\}\) is a collection of open sets in
\(X\) for
\(\alpha\) in some indexing set
\(I\) and let
\(O = \bigcup_{\alpha \in I} O_{\alpha}\text{.}\) By
Theorem 8.3, we know that
\(O_{\alpha}\) is a union of open balls for each
\(\alpha \in I\text{.}\) Combining all of these open balls together shows that
\(O\) is a union of open balls and is therefore an open set by
Theorem 8.3.
For part 2, assume that \(O_1\text{,}\) \(O_2\text{,}\) \(\ldots\text{,}\) \(O_n\) are open sets in \(X\) for some \(n \in \Z^+\text{.}\) To show that \(O = \bigcap_{k=1}^n O_k\) is an open set, we will show that \(O\) is a neighborhood of each of its points. Let \(x \in O\text{.}\) Then \(x \in O_k\) for each \(1 \leq k \leq n\text{.}\) Let \(k\) be between 1 and \(n\text{.}\) Since \(O_k\) is open, we know that \(O_k\) is a neighborhood of each of its points. So there exists \(\epsilon_k \gt 0\) such that \(B(x, \epsilon_k) \subseteq O_k\text{.}\) Since there are only finitely many values of \(k\text{,}\) let \(\epsilon = \min\{\epsilon_k \mid 1 \leq k \leq n\}\text{.}\) Then \(B(x, \epsilon) \subseteq B(x, \epsilon_k)\) for each \(k\) and so \(B(x, \epsilon) \subseteq \bigcap_{k=1}^n O_k = O\text{.}\) Therefore, \(O\) is a neighborhood of each of its points and \(O\) is an open set.
