(3.2.3) The Union of Open Sets is Open
Proof
Let \(\{O_{}: \lambda \in \Lambda\}\) be a collection of open sets and let \(O = \bigcup_{\lambda \in \Lambda}O_{\lambda}\). Let \(a\) be an arbitrary element of \(O\). To show that \(O\) is open, we need to produce an \(\epsilon\)-neighborhood of \(a\) that is contained in \(O\). Since \(a \in O\), then \(a\) must be in at least one particular set \(O_{\lambda'}\). But we know that \(O_{\lambda'}\) is open and so there must exists a neighborhood \(V_{\epsilon}(a) \subseteq O_{\lambda'}\). But since \(O_{\lambda'} \subseteq O\), then \(V_{\epsilon}(a) \subseteq O\) as we wanted to show.
For (ii), let \(\{O_1, O_2, ..., O_N\}\) be a finite collection of open sets. We know that \(a \in \bigcap_{k=1}^{N} O_k\). This means that there exists \(V_{\epsilon_k}(a) \subseteq O_{k}\). But since this is an intersection, the trick is to take the smallest one and therefore let \(\epsilon = min\{\epsilon_1, \epsilon_2, ..., \epsilon_N\}\). Since \(V_{\epsilon}(a)\) is the smallest one now, then we know \(V_{\epsilon}(a) \subseteq V_{\epsilon_k}(a)\) for all \(k\) and so \(V_{\epsilon}(a) \subseteq \bigcap_{k=1}^{N} O_k\) as we wanted to show. \(\blacksquare\).
Other Definitions and Properties
- Fo the definition of open, closed, compact, perfect,...etc sets see this.
- Fo the absolute value function definition and other properties, see this.
- For the definitions of sequences, subsequences and what it means to for a sequence to converge, see this.
- For the definitions of series, and what it means to for a series to converge, see this.
- For the "show the limit" template and an example, see this.
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