Point not in Subset of Metric Space iff Distance from Elements is Greater than Zero

Theorem

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

Let $H \subseteq A$ be an arbitrary subset of $A$.

Let $x \in A$ be arbitrary.

Then:

$x \notin H$

if and only if:

$\forall y \in H: \map d {x, y} > 0$

Proof

\(\ds \forall y \in H: \, \)

\(\ds \map d {x, y}\)

\(>\)

\(\ds 0\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \forall y \in H: \, \)

\(\ds \map d {x, y}\)

\(\ne\)

\(\ds \map d {x, x}\)

Metric Space Axiom $(\text M 1)$

\(\ds \leadstoandfrom \ \ \)

\(\ds \forall y \in H: \, \)

\(\ds y\)

\(\ne\)

\(\ds x\)

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\notin\)

\(\ds H\)

$\blacksquare$