Metric Space is T5

Theorem

Let $M = \struct {A, d}$ 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 $\map {B_{\epsilon_x} } x$ which is disjoint from $T$

each point $y \in T$ has an open $\epsilon$-ball $\map {B_{\epsilon_y} } y$ which is disjoint from $S$.

Then:

$U_S = \ds \bigcup_{x \mathop \in S} \map {B_{\epsilon_x / 2} } x$

$U_T = \ds \bigcup_{y \mathop \in T} \map {B_{\epsilon_y / 2} } y$

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

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

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces