I'm wondering if a connected countable metric space exists.

My intuition is telling me no.

For a space to be connected it must not be the union of 2 or more open disjoint sets.

For a set $M$ to be countable there must exist an injective function from $\mathbb{N} \rightarrow M$.

I know the Integers and Rationals clearly are not connected. Consider the set $\mathbb{R}$, if we eliminated a single irrational point then that would disconnect the set.

A similar problem arises if we consider $\mathbb{Q}^2$

In any dimension it seems by eliminating all the irrational numbers the set will become disconnected. And since $\mathbb{R}$ is uncountable there cannot exist a connected space that is countable.

My problem is formally proving this. Though a single Yes/No answer will suffice, I would like to know both the intuition and the proof behind this.

Thanks for any help.

I haven't looked at cofinite topologies (which I happened to see online). I also don't see where the Metric might affect the countability of a space, if we are primarily concerned with an injective function into the set alone.

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    $\begingroup$ I think your function from $\mathbb N \to M$ should be surjective for countability (I can find an injection from $\mathbb N\to \mathbb R$), and you also need to specify infinite otherwise a single point is connected and countable. $\endgroup$ – Mark Bennet Jan 2 '16 at 14:36
  • $\begingroup$ This is essentially a duplicate, in weaker form, of math.stackexchange.com/questions/1327483/… $\endgroup$ – Lee Mosher Jan 3 '16 at 22:14

Fix $x_0 \in X $. Then, the continuous(!) map $$ \Phi: X \to \Bbb {R}, x \mapsto d (x,x_0) $$ has an (at most) countable, connected image.

Thus, the image is a whole (nontrivial!, if $X $ has more than one point) interval, in contradiction to being countable.

EDIT: On a related note, this even show's that every connected metric space with more than one point has at least the cardinality of the continuum.

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    $\begingroup$ interesting answer. $\endgroup$ – Beginner Jan 2 '16 at 14:12
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    $\begingroup$ Very intuitive proof. You're using a result that says the only connected subsets of $\mathbb R$ are the intervals, I guess? $\endgroup$ – Jack M Jan 2 '16 at 21:47
  • $\begingroup$ @JackM: Yes, exactly. But that should be widely known. $\endgroup$ – PhoemueX Jan 2 '16 at 22:11

As usual, there is a trivial example that gives you no intuition: the set containing one point is trivially a metric space (in the only possible way).

If the space contains more than one point, see this answer for a proof of uncountability.


If by countable you include finite sets: yes, there exists. Take a single point.

However, if the space has more than one point and is connected, it must be uncountable. It is a simple result from Urysohn's Lemma combined with the "Intermediate Value Theorem". Note that Urysohn's Lemma is also a triviality on metric spaces.

Since it is a result from Urysohn's Lemma, this also holds on normal topological spaces. Moreover, it holds under the weaker hypothesis of regularity instead of normality (one can use the fact that a countable space is Lindelöf, and a regular Lindelöf space is normal).

An example that this does not hold on general topological spaces can be seen by taking the trivial topology on any uncountable set. However, there exists even connected Hausdorff spaces which are countably infinite (see this).


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