# Easy Metric Space Topology

#### Drexel28

MHF Hall of Honor
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
How about the metric space which contains only one point? (Giggle)

#### Drexel28

MHF Hall of Honor
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
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:
• Chris L T521 and Drexel28

#### Drexel28

MHF Hall of Honor
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.