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Show that if $E \subseteq X$ is connected and has at least two points, then $E$ has no isolated points. Thus a connected set with at least two points must contain infinitely many points.

I understand that if $E$ has no isolated points, it must be infinite, because no "ball" around the point will be such that it doesn't contain any element of E, but how do I write this mathematically?

I also understand that if $E$ is connected and has at least two points, it must contain infinitely many points because it is connected and so you must be able to have a "ball" around either point containing at least one element of $E$ by the definition of connectedness. I'm having trouble writing this mathematically and I feel I'm missing some intuition.

Here, $X$ is a nonempty set equipped with a metric $d$.

BCLC
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akeenlogician
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2 Answers2

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Hint Consider $a,b\in E$ consider $f:E\to\mathbb{R}$ defined by $f(x)=d(x,a)$ which is a continuous function , $f(a)=0,f(b)>0$ , $f(E)$ is connected and it is an interval.

Myshkin
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(Edit: after the edit this is no longer needed I think that as stated the result is false: take $X = \{a,b\}$ and put the indiscrete topology on it. It is connected (there are no separations of course) and it is not infinite.)

It is probably more interesting asking the same question with a $T1$ space: here points are closed. It can be proved that one point is isolated if and only if the singleton is open and hence you have that for connectdness you need to have no isolated points (the singleton would be both open and closed!)

  • Editted the question: X is a nonempty set equipped with the discreet metric, so I suppose the indiscreet topology doesn't work? – akeenlogician Oct 22 '13 at 04:52
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    No, of course it does not work now. Anyway in my answer there is still something useful for your problem since metric spaces are T2 and hence T1. –  Oct 22 '13 at 04:55