Every Metric Spaces is Hausdorff
Theorem
Every metric space is a Hausdorff space.
We will prove this by proving a more general and constructive result which will trivially imply the Hausforff property.
Lemma
If \(x, y \in X\) with \(x \neq y\), then if \(r_x + r_y \leq d(x, y)\),
\[ B_{r_x}(x) \cap B_{r_y}(y) = \varnothing.\]
Proof
We prove this by the contrapositive, that is assume that
\[ a \in B_{r_x}(x) \cap B_{r_y}(y).\]
Therefore
\[ d(a, x) < r_x \quad \text{and} \quad d(a, y) < r_y.\]
Now, summing these inequalities and applying the triangle inequality we have that
\[ d(x, y) \leq d(a, x) + d(a, y) < r_x + r_y.\]
Therefore we have that \(d(x, y) < r_x + r_y\), and therefore by the contrapositive
\[ r_x + r_y \leq d(x, y) \implies B_{r_x}(x) \cap B_{r_y}(y) = \varnothing.\]
Now we prove the Hausdorff property.
Proof
For any \(x, y \in X\) with \(x \neq y\) let \(r_x = r_y = \frac{d(x, y)}{2}\). As such, \(r_x + r_y = \frac{d(x, y)}{2} \leq \frac{d(x, y)}{2}\) as required to apply our lemma. Therefore
\[ B_{r_x}(x) \cap B_{r_y}(y) = \varnothing.\]