Distinct Points in Metric Space have Disjoint Neighborhoods

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Theorem

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

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


Then there exist neighborhoods $N_x$ and $N_y$ of $x$ and $y$ respectively such that $N_x \cap N_y = \O$, that is, that are disjoint.


Proof

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

From Distinct Points in Metric Space have Disjoint Open Balls, there exist disjoint open $\epsilon$-balls $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ containing $x$ and $y$ respectively.

From Open Ball is Neighborhood of all Points Inside it follows that $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ are neighborhoods of $x$ and $y$ respectively.

Hence the result.

$\blacksquare$


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