If $X, d$ is a connected metric with at least two points, prove it is uncountable.

Question

I was able to prove the theorem by contradiction, showing that if $X$ is a countable set, we can take $r_0 = min\{d(p, q) < r : \forall p, q \in X\}$ and so there exists an $N_r(p)$ such that it will not contain any point $q \in X : q \neq p$ in its neighborhood. Thus $p$ is not a limit point of $X$, nor is any other point. Then these $X$ is the union of singletons and since open, disjoint subsets of a metric are separated, so is $X$.

Is my proof entirely correct? Were there any incorrect terms used?

I was asking for proof-verification, not for an answer. I need to know if my proof is right.

Please clarify the question. What is the topology on $X$. We cannot see your book. — , Nov 05 '13 at 00:26
Your definition of $r_0$ is a little confusing. Where'd the $r$ come from, for example? — zibadawa timmy, Nov 05 '13 at 01:00
$r_0$ is the smallest distance between any two points in the metric space. — Don Larynx, Nov 05 '13 at 01:12
.... If your space has only two points, why would it be uncountable? — , Nov 05 '13 at 02:47
Your $r_0$ is not guaranteed to be nonzero. Consider the rationals. — Cameron L. Williams, Nov 05 '13 at 04:35
Please search a bit before asking a question. There have been 194,737 questions already on this site: there is an immense chance that yours has already been asked, so it is a good way to minimize effort (yours and that of other people!) to make reasonably sure that you are not going to duplicate an old question. — Mariano Suárez-Álvarez, Nov 05 '13 at 04:47
If I google «connected metric space uncountable» the very first result is precisely the old question. If you want extra information about anyone's old answers, use comments to ask for them, and avoid «what the heck» and other expressions which certainly do not motivate anyone to help you. — Mariano Suárez-Álvarez, Nov 05 '13 at 04:51

score 1 · Accepted Answer · answered Nov 05 '13 at 04:44

I think the quickest solution is as follows. If $X$ is connected, then so is $X\times X$. Thus, $d(X\times X)\subseteq\mathbb{R}$ is connected, since $d$ is continuous. But, if $X$ is not a point, then we know that $d(X\times X)$ contains more than just $0$. But, since $d(X\times X)$ is an interval, we are done since any interval with more than one point is uncountable.

I don't know if this counts as "constructive".

If $X, d$ is a connected metric with at least two points, prove it is uncountable.

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