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 $B_\epsilon(x) \cap B_\delta(y) \neq \varnothing$

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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

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Is there a more satisfying solution i.e. closed form expression for reduced $\epsilon, \delta$ so they are no longer intersecting. Thanks!

  • $\begingroup$ Choosing two parameter is too difficult. Hausdorff condition does not require the arbitrariness. Let $\varepsilon = d(x,y)$. Choose $r=\varepsilon/3$. Then B_r (x) \cap B_r (y) =\varnothing$. $\endgroup$ – Will Kwon Jul 17 '16 at 22:27

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.

  • $\begingroup$ Okay I see, so convenient radius basically $\endgroup$ – Carlos - the Mongoose - Danger Jul 17 '16 at 22:31
  • $\begingroup$ Yes, you have to find at least one radius that works. Anything less than $d(x,y)/2$ will do. $\endgroup$ – Ivo Terek Jul 17 '16 at 22:32
  • $\begingroup$ Can you provide a hint for that triangle inequality part? Suppose $\exists z \in B(x, d/2) \cap B(y, d/2)$, then $d(x,z) \leq d(x,y) + d(y,z)$, $d(x,z) \leq d(x,y)/2$, $d(y,z) \leq d(x,y)/2$ $\implies$ $d(x,z) + d(y,z) \leq d(x,y)$... Im not seeing a contradiction yet. Actually the last line might be a contradiction since I have $d(x,z) + d(y,z) = d(x,z) + d(z,y) \leq d(x,y)$ which cannot happen $\endgroup$ – Carlos - the Mongoose - Danger Jul 17 '16 at 22:41
  • $\begingroup$ The idea is to take $z$ in the intersection, and estimate $d(x,y)$ using the triangle inequality. Then use that $z$ is in both balls to get $d(x,y) < d(x,y)$, which is not possible. Can you write it? $\endgroup$ – Ivo Terek Jul 17 '16 at 22:46
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    $\begingroup$ Something like $d(x,y) \leq d(x,z) + d(z,y)$. But $d(x,z) < d/2$ and $d(y,z) < d/2$, so we have $d$ < d(x,z) + d(y,z) < $d$. A contradiction $\endgroup$ – Carlos - the Mongoose - Danger Jul 17 '16 at 22:52

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