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Math Help - Easy Metric Space Topology

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    MHF Contributor Drexel28's Avatar
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    Easy Metric Space Topology

    Problem: Does there exist a countable metric space \left(\mathcal{M},d\right) which is connected? If so, give an example and prove it's in fact connected.
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    MHF Contributor Bruno J.'s Avatar
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    How about the metric space which contains only one point?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    How about the metric space which contains only one point?
    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.
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    Choose x_0\in \mathcal{M}. If it is an isolated point then \{x_0\} is open and closed, so \mathcal{M} is not connected. Otherwise, let D = \{d(x,x_0):x\in\mathcal{M},\;x\ne x_0\}. Then D is a countable set of positive real numbers, with \inf D = 0. It may happen that D is not bounded above, in which case define \sup D = \infty. In any case, we can find a real number r\notin D with 0<r<\sup D. Let U = \{x\in\mathcal{M}:d(x,x_0)<r\}, V = \{x\in\mathcal{M}:d(x,x_0)>r\}. Then U,\;V are disjoint nonempty open subsets of \mathcal{M} with U\cup V =\mathcal{M}. So \mathcal{M} cannot be connected.

    It is essential for that argument that \mathcal{M} should be a metric space. This example shows that a countable Hausdorff topological space can be connected.
    Last edited by Opalg; May 19th 2010 at 01:42 PM. Reason: added the word "nonempty"
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Opalg View Post
    Choose x_0\in \mathcal{M}. If it is an isolated point then \{x_0\} is open and closed, so \mathcal{M} is not connected. Otherwise, let D = \{d(x,x_0):x\in\mathcal{M},\;x\ne x_0\}. Then D is a countable set of positive real numbers, with \inf D = 0. It may happen that D is not bounded above, in which case define \sup D = \infty. In any case, we can find a real number r\notin D with 0<r<\sup D. Let U = \{x\in\mathcal{M}:d(x,x_0)<r\}, V = \{x\in\mathcal{M}:d(x,x_0)>r\}. Then U,\;V are disjoint nonempty open subsets of \mathcal{M} with U\cup V =\mathcal{M}. So \mathcal{M} cannot be connected.

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

    Mine:

    Spoiler:
    My argument was that if \mathcal{M} is countable so is \mathcal{M}\times\mathcal{M} and thus \text{dist}\left(\mathcal{M}\times\mathcal{M}\righ  t). So, let x_0,y_0\in\mathcal{M} be distinct (this is by assumption since the metric space has more than one point). Let

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

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

    \mathcal{M} is not connected


    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 \mathcal{M} is a countable metric space without an isolated point then \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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