# Metric topology problem

## 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

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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?

member 587159
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.

• facenian