Oh, I forgot to mention I use the term countable to mean countably infinite. But, the problem can be extended to say any at most countable metric space with more than one point.How about the metric space which contains only one point? (Giggle)

Choose \(\displaystyle x_0\in \mathcal{M}\). If it is an isolated point then \(\displaystyle \{x_0\}\) is open and closed, so \(\displaystyle \mathcal{M}\) is not connected. Otherwise, let \(\displaystyle D = \{d(x,x_0):x\in\mathcal{M},\;x\ne x_0\}\). Then D is a countable set of positive real numbers, with \(\displaystyle \inf D = 0\). It may happen that D is not bounded above, in which case define \(\displaystyle \sup D = \infty.\) In any case, we can find a real number \(\displaystyle r\notin D\) with \(\displaystyle 0<r<\sup D.\) Let \(\displaystyle U = \{x\in\mathcal{M}:d(x,x_0)<r\}\), \(\displaystyle V = \{x\in\mathcal{M}:d(x,x_0)>r\}\). Then \(\displaystyle U,\;V\) are disjoint nonempty open subsets of \(\displaystyle \mathcal{M}\) with \(\displaystyle U\cup V =\mathcal{M}\). So \(\displaystyle \mathcal{M}\) cannot be connected.

It is essential for that argument that \(\displaystyle \mathcal{M}\) should be a metric space. This example shows that a countable Hausdorff topological space can be connected.

It is essential for that argument that \(\displaystyle \mathcal{M}\) should be a metric space. This example shows that a countable Hausdorff topological space can be connected.

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Nice!Choose \(\displaystyle x_0\in \mathcal{M}\). If it is an isolated point then \(\displaystyle \{x_0\}\) is open and closed, so \(\displaystyle \mathcal{M}\) is not connected. Otherwise, let \(\displaystyle D = \{d(x,x_0):x\in\mathcal{M},\;x\ne x_0\}\). Then D is a countable set of positive real numbers, with \(\displaystyle \inf D = 0\). It may happen that D is not bounded above, in which case define \(\displaystyle \sup D = \infty.\) In any case, we can find a real number \(\displaystyle r\notin D\) with \(\displaystyle 0<r<\sup D.\) Let \(\displaystyle U = \{x\in\mathcal{M}:d(x,x_0)<r\}\), \(\displaystyle V = \{x\in\mathcal{M}:d(x,x_0)>r\}\). Then \(\displaystyle U,\;V\) are disjoint nonempty open subsets of \(\displaystyle \mathcal{M}\) with \(\displaystyle U\cup V =\mathcal{M}\). So \(\displaystyle \mathcal{M}\) cannot be connected.

It is essential for that argument that \(\displaystyle \mathcal{M}\) should be a metric space. This example shows that a countable Hausdorff topological space can be connected.

Mine:

\(\displaystyle \text{dist}(x_0,y_0)=\xi>0\). Then, since \(\displaystyle (0,\xi)\subseteq\mathbb{R}\) is uncountable there is some \(\displaystyle \delta\in(0,\xi)-\text{dist }\left(\mathcal{M}\times\mathcal{M}\right)\). So, noticing that \(\displaystyle \varphi:\mathcal{M}\to\mathcal{M}:x\mapsto \text{dist}(x,x_0)\) is continuous the sets \(\displaystyle \varphi^{-1}(-\infty,\delta)\)

and \(\displaystyle \varphi^{-1}(\delta,\infty)\) are open. But, they are also disjoint and non-empty \(\displaystyle x_0\in\varphi^{-1}(-\infty,\delta),y_0\in\varphi^{-1}(\delta,\infty)\). And by choice of \(\displaystyle \delta\) we have that \(\displaystyle \varphi^{-1}(-\infty,\delta)\cup\varphi^{-1}(\delta,\infty)=\mathcal{M}\) which proves that

\(\displaystyle \mathcal{M}\) is

And while I was aware of the other result you quoted there is something that can be said even more specifically for countable metric spaces. Namely if \(\displaystyle \mathcal{M}\) is a countable metric space without an isolated point then \(\displaystyle \mathcal{M}\approx \mathbb{Q}\) with the obvious metric. From there the conclusion would follow. It's a nice proof too.

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