Metric topology

Metrizable implies Hausdorff

Let (𝑋,𝑑) be a metric space. Then 𝑋 is Hausdorff under its metric topology. topology

Proof

Let 𝑥,𝑦 ∈𝑋 with 𝑥 ≠𝑦 (if no such 𝑥,𝑦 exist then the conclusion is trivially satisfied). Let 𝑟 =𝑑(𝑥,𝑦). Then B𝑟/2(𝑥) and B𝑟/2(𝑦) are disjoint open neighbourhoods of 𝑥 and 𝑦 respectively. Since if 𝑧 ∈B𝑟/2(𝑥) ∩B𝑟/2(𝑦) then 𝑑(𝑥,𝑦) ≤𝑑(𝑥,𝑧) +𝑑(𝑦,𝑧) <𝑟 which is a contradiction.

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