Connected metric spaces with at least 2 points are uncountable.

Question

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?

Answer 1

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 :)

Answer 2

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.

Answer 3

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.