Easy Metric Space Topology

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Problem: Does there exist a countable metric space \(\displaystyle \left(\mathcal{M},d\right)\) which is connected? If so, give an example and prove it's in fact connected.
 

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
How about the metric space which contains only one point? (Giggle)
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
How about the metric space which contains only one point? (Giggle)
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.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
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.
 
Last edited:

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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.
Nice!

Mine:

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

\(\displaystyle \text{dist}(x_0,y_0)=\xi>0\). 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 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 \(\displaystyle \mathcal{M}\) 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.