Metrizable implies Hausdorff

Let $(𝑋, 𝑑)$ be a metric space. Then $𝑋$ is Hausdorff under its metric topology. #m/thm/topology

Proof

Let $𝑥, 𝑦 \in 𝑋$ with $𝑥 \neq 𝑦$ (if no such $𝑥, 𝑦$ exist then the conclusion is trivially satisfied). Let $𝑟 = 𝑑 (𝑥, 𝑦)$ . Then $B 𝑟 / 2 (𝑥)$ and $B 𝑟 / 2 (𝑦)$ are disjoint open neighbourhoods of $𝑥$ and $𝑦$ respectively. Since if $𝑧 \in B 𝑟 / 2 (𝑥) \cap B 𝑟 / 2 (𝑦)$ then $𝑑 (𝑥, 𝑦) \leq 𝑑 (𝑥, 𝑧) + 𝑑 (𝑦, 𝑧) < 𝑟$ which is a contradiction.

#state/tidy | #lang/en | #SemBr