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Let $X := \{{1, 2, 3}\}$ and $τ := \{{∅, \{{1}\}, \{{1, 2}\}, \{{1, 3}\}, X}\}$ be a topology space. Is it metrizable?

I am learning about topological space.

I read about metrizable topology space in Wikipedia and this post There is no difference between a metrizable space and a metric space (proof included)..

I have NO idea how to approach this problem , any help is welcome.

Thanks !

Kevin.S
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Algo
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    Metrizable spaces are Hausdorff. – Randall Nov 16 '22 at 19:42
  • I didn't learn about Hausdorff ,so I have to solve the problem another way. – Algo Nov 16 '22 at 19:50
  • Then all you need is the triangle inequality. – Randall Nov 16 '22 at 19:55
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    You don't even need the triangle inequality. Let $a$ be the distance between points $1$ and $2$. By definition of metric, $a>0$. The open ball of radius $a$ about $2$ is an open set. Figure out what this set can be, and obtain a contradiction to the given $\tau$. – Andreas Blass Nov 16 '22 at 20:00

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No. A topology $\tau$ on a finite set $X$ is metrizable if and only if $\tau$ is discrete. Your $X$ does not have the discrete topology.

  1. The discrete topology on any set $X$ is metrizable. Simply take $d(x,x) = 0$ and $d(x,y)=1$ for $x \ne y$.

  2. If $d$ is a metric on the finite set $X$, then the metric topology is discrete. To see this, let $r = \min\{d(x,y ) \mid x,y \in Y, x \ne y \}$. Then $r > 0$ and for each $x \in X$ we have $U_r(x) = \{ y \in X \mid d(x,y) < r\} = \{x\}$.

Paul Frost
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