Open Balls of Same Radius form Open Cover

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Therorem

Let $M = \struct{A, d}$ be a metric space.

Let $\UU_\epsilon = \set{\map {B_\epsilon} x : x \in A}$

That is, $\UU_\epsilon$ is the set of all open balls of radius $\epsilon > 0$.

Then:

$\UU_\epsilon$ is an open cover of $M$.

Proof

From Open Ball is Open Set:

$\UU_\epsilon$ is a set of open subsets

From Center is Element of Open Ball:

$\forall x \in A : x \in \map {B_\epsilon} x$

By definition, $\UU_\epsilon$ is a cover of $A$.

By definition, $\UU_\epsilon$ is an open cover of $A$.

$\blacksquare$