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Prove that a metric space with the discrete metric and more than one element is not connected.

Attempt:Let $X$ be a discrete metric space with more than one element. If $A$ is any nonempty proper subset of $X$, $A$ and $X-A$ is a separation of $X$, since all subsets of discrete metric spaces are open.

I realize this proof is the same proof that spaces with the discrete topology are not connected, is it appropriate to prove this result for metric spaces like this?

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Connectedness is a topological property, it does not really depend on the metric, just on its induced topology (defined in terms of open and closed sets etc.). The discrete metric induces the discrete topology.

So I'd say it's totally appropriate. Just let it stand.

Henno Brandsma
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