# Connected metric spaces with at least 2 points are uncountable.

That's a problem I proved (quite a while back) in tiny Rudin. However, I don't really get it. The other questions were actually useful results - I don't think I've ever come near using this result. Surely it's going to be close to apparent that you're working in an uncountable set?

For instance, examples where this result could be applied but it is hard otherwise to tell that the space is uncountable?

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I edited the title because the old one sounded to me like you were talking about metric spaces with 2 connected points. –  Alex Becker Jan 5 '12 at 2:24
@JonasMeyer "Are you asking for examples where this result could be applied but it is hard otherwise to tell that the space is uncountable?" This is how I interpret the question. –  Alex Becker Jan 5 '12 at 2:49
Well, for instance this leads to a proof of the uncountability of $\mathbb{R}$ (not necessarily the easiest, but a nice enough one, I think). In general I think I agree that this result is not something you are going to use to prove some other result: neither is Fermat's Last Theorem, by the way. Not all results in mathematics have to be directly useful to be interesting. –  Pete L. Clark Jan 5 '12 at 3:24
The result says that aside from a trivial one-point topology, there is no countable connected metric space. The "application" of this theorem is not to prove that a set is uncountable, but to tell everyone to stop looking for countable connected metric spaces. –  Austin Mohr Jan 5 '12 at 3:35
Just a small addition: there are countable connected spaces (e.g. the integers with the cofinite topology). There are even countable connected Hausdorff spaces (the first example of which is due to Urysohn). See also this MO thread. –  t.b. Jan 5 '12 at 4:09

Let your points be $a$ and $b$.

Let $\lambda\in(0,1)$ Suppose there is no point $x$ in your space such that $d(a,x)=\lambda d(a,b)$. Then the sets $\{z:d(a,z)<\lambda d(a,b)\}$ and $\{z:d(a,z)>\lambda d(a,b)\}$ are two non-empty open sets which partition the space.

Since we are assuming connectedness, this is impossible.

Therefore, the image of the function $d(a,\mathord\cdot):X\to\mathbb R$ contains the interval $(0,d(a,b))$, and this can only happen if $X$ is uncountable.

Even if useless, this is a pretty result :)

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The question wasn't asking to prove the result. –  Dustan Levenstein Jan 5 '12 at 2:32
The OP seems to indicate that he has already proved it, but is now looking for "non-trivial examples." I am not sure what the question is. –  Jonas Meyer Jan 5 '12 at 2:33
I do not wish to +1 as I believe it would be done for the sake of me going against the downvotes, and +'s should only be used when genuinely that user has made a good comment. However, I don't think the post warrants downvotes as it's hardly fair to say you've never posted an answer without entirely reading a question (or maybe you have never, in which case you should give yourself a pat on the back). –  Adam Jan 5 '12 at 3:18
Alos, this isn't a comment at anybody in particular I just think people are very fast at -voting good hearted (?) mistakes. –  Adam Jan 5 '12 at 3:19
@Adam: Downvoting an answer isn't a moral judgement on the answerer. It is a judgement that the answer is not a good answer. –  Chris Eagle Jan 5 '12 at 19:44

There may be situations I am unaware of, but I don't think the standard setting is to have a metric space $(X,d)$, where you know $X$ is connected under $d$, and that there are at least two distinct points in $X$, but don't already know the space is uncountable (and care!). I think it would be far more likely to know that $X$ is at most countable, and then we would know the space must be disconnected, regardless of the metric we choose to use.

For example, for those that haven't seen Ostrowski's theorem, and have no idea what metrics can be placed on $\mathbb{Q}$, your result immediately shows it is impossible to construct a metric under which $\mathbb{Q}$ is a connected metric space. (That's not to say it's a bad idea to get your hands dirty, try to build a metric $d'$ so that $(\mathbb{Q},d')$ is a connected metric space, and see what goes wrong!)

One could then see this as an argument to construct $\mathbb{R}$ from $\mathbb{Q}$, since no matter what metric we use, there are holes.

I suppose one could also say, if there is a topology $\tau$ on $\mathbb{Q}$ so that $(\mathbb{Q},\tau)$ is a connected topological space, then we also know that this space is not metrizable. I don't know if this is a particularly useful point of view though..

Of course these are only examples using $\mathbb{Q}$ to illustrate the point, and the same holds for far more odd 'looking' at most countable spaces, where it may be far less intuitive that there are no metrics to make the space a connected metric space.

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Call those two points $x_1$ and $x_2$. So $d(x_1,x_2)>0$.

What is the set $\{d(x_1,x) : x\in A\}$? Does it contain all numbers between $0$ and $d(x_1,x_2)$? If so, then you have at least as many points $x\in A$ as numbers between $0$ and $d(x_1,x_2)$. If not, then some number $c$ between $0$ and $d(x_1,x_2)$ is not the distance between any point $x\in A$ and $x_1$. So consider the two sets $$\{x\in A : d(x_1,x)<c\}\text{ and }\{x\in A : d(x_1,x)>c\}.$$ Those are disjoint open subsets of $A$ whose union is all of $A$, so $A$ would not be connected.

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@MichaelHardy: I am the only downvote on this answer, and I haven't downvoted anything else in the last two weeks, apart from another answer to this question. (I downvoted this answer because the question was not asking for a proof) –  Ben Millwood Jul 18 '14 at 12:03

The biggest application that I know for this result is the immediate corollary that a countable metric space is totally disconnected.

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Well, I don't have any interesting examples of connected metric spaces which are not immediately obviously uncountable, but here's a stronger theorem:

Every countable regular $T_1$ with at least two points is disconnected.

There are a couple of ways to prove this that I know of; a separation of two arbitrary points can be built successively by taking bigger and bigger disjoint open sets around them with disjoint closures. Alternatively, it can be shown that such a space is necessarily normal, and then Urysohn's lemma can be used to show that any two points are separated.

It is tricky to come up with an example of a countably infinite space which is regular and $T_1$, but not metric (ie, a space which confirms that this theorem really is stronger than the one you gave). I'll withhold my examples for now, in case you decide you want to tackle that problem yourself. As t.b. mentioned in the comments, there are countable Hausdorff connected spaces; that is a more difficult problem to figure out, if you haven't already followed the link.

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A mildly interesting remark:

A comment suggests this result can be used to prove the uncountability of $R$. But the proofs here and elsewhere of this result typically use the existence of a surjective function to an interval $[a,b]$ and the uncountability of that interval. So the argument suggested in the comment would be circular. Here, however, is a nice proof of this result that does not use uncountability of $R$.

Let $X = \{x_n\}_{n=1}^\infty$ be a connected metric space and suppose $\epsilon >0$. Let $U_n$ be the open ball of radius $\epsilon/2^n$ centered at $x_n$. Define a relation ~ on $X$ by $x$ ~ $y$ if there is a sequence ${n_i}$, $1 \le i \le k$ such that $x \in U_{n_1}$, $y \in U_{n_k}$ and $U_{n_i} \cap U_{n_{i+1}} \ne \emptyset.$ Let $z_{n_i}$ be a point in $U_{n_i} \cap U_{n_{i+1}}$

It is immediate that ~ is an equivalence relation and that equivalence classes are open sets. Since $X$ is connected there can be only one equivalence class. It is also easy to see that $x$ ~ $y$ implies $$d(x,y) < 2\sum_{i=1}^{k-1}d(x_{n_i}, x_{n_{i+1}}) < 2\sum_{i=1}^{k-1}(d(x_{n_i}, z_{n_{i}}) + d(z_{n_i}, x_{n_{i+1}}))<4\sum_{i=1}^k \frac{\epsilon}{2^{n_i}} < 4\epsilon.$$ Since $x,y$ and $\epsilon >0$ are arbitrary, it follows that the diameter of $X$ is $0$ and $X$ is a singleton.

Another "easy" proof of uncountability of $R$ uses the fact that the Lebesgue measure of an interval is positive and countable sets have measure zero. But the common proof of fact that the measure of a non-trivial interval is positive uses connectivity and is similar to the argument above. I think if one tries to dig into the topological roots of the measure theory proof you end up with something like the argument above.

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