Topology: An Inquiry-Based Approach

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 $O_1\text{,}$ $O_2\text{,}$ $\dots \text{,}$ $O_n$ are open sets in $X$ for some $n\in {\mathbb{Z}}^{+}\text{.}$ 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 $1\le k\le n\text{.}$ 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 ${ϵ}_k>0$ such that $B(x,{ϵ}_k)\subseteq O_k\text{.}$ Since there are only finitely many values of $k\text{,}$ let $ϵ=min\{{ϵ}_k\mid 1\le k\le n\}\text{.}$ Then $B(x,ϵ)\subseteq B(x,{ϵ}_k)$ for each $k$ and so $B(x,ϵ)\subseteq \bigcap _{k=1}^n{O}_k=O\text{.}$ Therefore, $O$ is a neighborhood of each of its points and $O$ is an open set.