# Easy Metric Space Topology

• May 17th 2010, 08:14 PM
Drexel28
Easy Metric Space Topology
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.
• May 18th 2010, 01:45 PM
Bruno J.
How about the metric space which contains only one point? (Giggle)
• May 18th 2010, 04:23 PM
Drexel28
Quote:

Originally Posted by Bruno J.
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.
• May 19th 2010, 10:52 AM
Opalg
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 PM
Drexel28
Quote:

Originally Posted by Opalg
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:

Spoiler:
My argument was that if $\displaystyle \mathcal{M}$ is countable so is $\displaystyle \mathcal{M}\times\mathcal{M}$ and thus $\displaystyle \text{dist}\left(\mathcal{M}\times\mathcal{M}\righ t)$. 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.