Can a connected metric space with at least two points be countable?

[SiddharthM]

I was doing a Rudin problem and wanted to know if there was a cleaner way of going about this - maybe a direct one.

So the question asks to prove that every connected metric space with at least two points is uncountable. The hint given in Rudin uses the thm below

Fix p in X, delta greater than zero, define A to be the delta neighborhood of p and B to be all x in X s.t. d(p,x)>delta. Then A and B are separated.

The show that a connected metric space with at least 2 points cannot be countable I show that the function f: X-->[0,infinity), with f(x)=d(p,x)

cannot be onto under the assumption that X is countable.* Hence there is a positive delta I can apply the above theorem (if the range of X under f is bounded then use supf(X)=alpha and apply the same idea to the function f:X-->[0,alpha)..that is show that the function cannot be onto) to show X is disconnected to give a contradiction.

* The argument here is assume that it is onto, then for each x in the nonnegative reals inversef(x) we get a nonempty subset of X. If x and y are distinct nonnegative reals then it's easy to see that inversef(x) and inversef(y) are disjoint sets. Then inversef([0,infinity) ) is an uncountable union of nonempty disjoint sets and is therefore uncountable and a subset of X which yields a contradiction. Hence there is a delta in [0,infinity) s.t. inversef(delta) is empty.

It's a bit lumpy with all the proof-by-contradictions. Btw, I'm ommitting the inductive proof that shows X cannot be finite.

 

[fresh_42]

The way you ruled out the finite case should work with the countable infinite case as well.

If we had path connection, then we could transport the connectedness of an interval by a continuity argument onto $X$. But path connection is stronger than connection. E.g.
$$
X=\{\,(0,y)\,|\,-1\leq y\leq 1\,\}\cup \{\,(x,\sin \frac{1}{x})\,\,|\,x\in \mathbb{R}-\{\,0\,\}\}
$$
is connected but not path connected.