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A Set is Open in the Metric Topology iff it contains a Ball around each Point

Last updated Nov 1, 2022

# Statement

Let $(M, d)$ be a Metric Space endowed with the Metric Topology. $U \subset M$ is Open If and Only If $\forall x \in U$, there exists $\epsilon > 0$ s.t. $B_{\epsilon}(x) \subset U$.

# Proof

($\Rightarrow$): Since we are in the Metric Topology, there exists Index Set $I$ and ${B_{\epsilon_\alpha}(x_\alpha)}{\alpha \in I}$ so that $$U = \bigcup\limits{\alpha \in I} B_{\epsilon_\alpha}(x_\alpha)$$ Let $y \in U$. Then there exists $\alpha \in I$ so that $y \in B_{\epsilon_\alpha}(x_\alpha)$. Now apply Finite Intersections of Open Balls contain Open Balls about each point to $B_{\epsilon_\alpha}(x_\alpha) \cap B_{\epsilon_\alpha}(x_\alpha)$ to get $\zeta > 0$ so that $B_{\zeta}(y) \subset B_{\epsilon_\alpha}(x_{\alpha}) \subset U$. $\checkmark$

$(\Leftarrow)$: Let $\epsilon_{x} > 0$ be such that $B_{\epsilon_{x}}(x) \subset U$ for all $x \in U$. Then $\bigcup\limits_{x \in U}B_{\epsilon_{x}}(x) = U$. Since $B_{\epsilon_{x}}(x)$ is Open for each $x \in U$, $U$ is Open. $\checkmark$ $\blacksquare$

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