Does there exist a countable metric space $\displaystyle \left(\mathcal{M},d\right)$ which is connected? If so, give an example andProblem:proveit's in fact connected.

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- May 17th 2010, 08:14 PMDrexel28Easy Metric Space Topology
Does there exist a countable metric space $\displaystyle \left(\mathcal{M},d\right)$ which is connected? If so, give an example and__Problem:____prove__it's in fact connected. - May 18th 2010, 01:45 PMBruno J.
How about the metric space which contains only one point? (Giggle)

- May 18th 2010, 04:23 PMDrexel28
- May 19th 2010, 10:52 AMOpalg
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. - May 19th 2010, 12:52 PMDrexel28
Nice!

Mine:

__Spoiler__:

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.