Metric topology problem

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Homework Statement


This is a problem from Munkres(Topology): Show that a connected metric space ##M## having having more than one point is uncountable.

Homework Equations


A theorem of that section of the book states: Let ##X## be a nonempty compact Hausdorff space. If no singleton in ##X## is open, then ##X## is uncountable


The Attempt at a Solution

d[/B]
If ##M## is connected and has more than one point then no singleton can be an open set since ##\{x\}\,\text{and}\,X\setminus\{x\}## would be a separation. Then if ##M## were compact application of the above mentioned theorem shows that ##M## is uncountable.
However the problem only states that ##M## is connected.
Is it possible that that every connected metric space be compact? or the problem should be solved without using the above mentioned theorem?
 

Answers and Replies

  • #2
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No. Certainly not every connected metric space is compact. Take for example ##\mathbb{R}## with the usual topology.

I don't see a way to use the theorem you listed. Here's a hint for a proof:

Let ##x,y## be 2 distinct points in the metric space ##M##. Consider the set ##S :=\{d(x,z): z \in M\}##.

You have to distinguish 2 cases:

(1) S contains the interval ##[0,d(x,y)]##
(2) S does not contain the interval ##[0,d(x,y)]##

The second will lead to a contradiction as ##M## is connected, and the first shows you that ##S## must be uncountable, as it contains an interval.
 
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