Distinct Points in Metric Space have Disjoint Neighborhoods

Theorem

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

Let $x, y \in M: x \ne y$.

Then there exist disjoint $\epsilon$-neighborhoods $N_\epsilon \left({x}\right)$ and $N_\epsilon \left({y}\right)$ containing $x$ and $y$ respectively.

Proof

Let $x, y \in A: x \ne y$.

Then $d \left({x, y}\right) > 0$.

Put $\epsilon = \dfrac {d \left({x, y}\right)} 2$.

Let $N_\epsilon \left({x}\right)$ and $N_\epsilon \left({y}\right)$ be the $\epsilon$-neighborhoods of $x$ and $y$ respectively.

Suppose $N_\epsilon \left({x}\right)$ and $N_\epsilon \left({y}\right)$ are not disjoint.

Then $\exists z \in M$ such that $z \in N_\epsilon \left({x}\right)$ and $z \in N_\epsilon \left({y}\right)$.

Then $d \left({x, z}\right) < \epsilon$ and $d \left({z, y}\right) < \epsilon$.

Hence $d \left({x, z}\right) + d \left({z, y}\right) < 2 \epsilon = d \left({x, y}\right)$.

This contradicts the definition of a metric, so there can be no such $z$.

Hence the neighborhoods must be disjoint.

$\blacksquare$