I wish to show that any metrizable space $(X,\mathcal{T})$ is Hausdorff
Proof attempt:
Let $d$ be the metric that generates the topology on $X$. Pick two points $x,y \in X$, we wish to produce two disjoint open sets that separates $x,y$.
Let $B_\epsilon(x)$ and $B_\delta(y)$ be two metric balls containing $x,y$ respectively. Suppose that $B_\epsilon(x) \cap B_\delta(y) \neq \varnothing$
Stuck Here: Hmm…How should I adjust $\epsilon, \delta$ so that these
balls are separated?
Idea: Reduce $\epsilon$ by half. If they are still intersecting…reduce $\delta$ by half. Continue ad infinitum
Is there a more satisfying solution i.e. closed form expression for
reduced $\epsilon, \delta$ so they are no longer intersecting. Thanks!
Best Answer
Hint: It essentialy boils down to showing that metric spaces are Hausdorff. Take $x,y \in X$ with $x \neq y$. Then $d(x,y) > 0$. Can you check that $$B\left(x,\frac{d(x,y)}{2} \right) \cap B\left(y,\frac{d(x,y)}{2}\right) = \varnothing?$$You should convince yourself with a drawing. To prove it formally, take $z$ in that intersection and get a contradiction using the triangle inequality.