I was doing a Rudin problem and wanted to know if there was a cleaner way of going about this - maybe a direct one.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Connected Metric Spaces

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