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If $|X|<\infty$ then T metrisable $\rightarrow$ T discrete topology.

I said let $d$ be a metric and let $x \in X$. I want to show that $\{x\}$ is open. How do I show this?

Paul
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2 Answers2

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Let $r$ be the infimum of all $d(x,y)$ with $y\ne x$. As we take the infimum over a finite set, it is in fact a minimum. As all $d(x,y)$ are positive, so is the minimum. $B_r(x)=\{x\}$.

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Here's a fun proof that uses a 'sledgehammer' theorem.

Let $\tilde{X}$ be the topological space which has the same elements as $X$, and whose topology is discrete. Let $f\colon \tilde{X} \to X$ be given by $f(x)=x$. The function $f$ is clearly continuous as $\tilde{X}$ is discrete, and is a bijection. Recall that a metrisable space is Hausdorff and, as $X$ is finite, $\tilde{X}$ is compact. Hence, $f$ is a continuous bijection from a compact space to a Hausdorff space and so is a homeomorphism. As $\tilde{X}$ is discrete, it follows that any space homeomorphic to $\tilde{X}$ is also discrete.

Dan Rust
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