Metric Space is T5

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Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Then $M$ is a $T_5$ space.


Proof

Let $S, T \subseteq A$ such that $S$ and $T$ are separated in $A$.

Then:

each point $x \in S$ has an open $\epsilon$-ball $B_{\epsilon_x} \left({x}\right)$ which is disjoint from $T$
each point $y \in T$ has an open $\epsilon$-ball $B_{\epsilon_y} \left({y}\right)$ which is disjoint from $S$.

Then:

$U_S = \displaystyle \bigcup_{x \mathop \in S} B_{\epsilon_x / 2} \left({x}\right)$
$U_T = \displaystyle \bigcup_{y \mathop \in T} B_{\epsilon_y / 2} \left({y}\right)$

are disjoint open neighborhoods of $S$ and $T$ respectively.

Hence the result by the definition of $T_5$ space.

$\blacksquare$


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