Is Every Connected Metric Space Compact?

[facenian]

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.