a connected normal space with more than one point is uncountable

The proof of the following result is an application of the generalized intermediate value theorem (along with Urysohn’s lemma):

Proposition.

A connected normal space with more than one point is uncountable.

Proof.

Let $X$ be a http://planetmath.org/node/941connected http://planetmath.org/node/1532normal space with at least two distinct points $x_1$ and $x_2$. As the sets $\{x_1\}$ and $\{x_2\}$ are http://planetmath.org/node/2739closed and disjoint, Urysohn’s lemma furnishes a continuous function $f:X\to [0,1]$ such that $f(x_1)=0$ and $f(x_2)=1$. Because $X$ is connected, the generalized intermediate value theorem implies that $f$ is surjective. Thus $f$ may be suitably to give a bijection between a subset of $X$ and the uncountable set $[0,1]$, from which it follows that $X$ is uncountable. ∎

Title	a connected normal space with more than one point is uncountable
Canonical name	AConnectedNormalSpaceWithMoreThanOnePointIsUncountable
Date of creation	2013-03-22 17:17:46
Last modified on	2013-03-22 17:17:46
Owner	azdbacks4234 (14155)
Last modified by	azdbacks4234 (14155)
Numerical id	10
Author	azdbacks4234 (14155)
Entry type	Theorem
Classification	msc 54D05
Related topic	UrysohnsLemma
Related topic	NormalTopologicalSpace
Related topic	Uncountable
Related topic	Bijection
Related topic	ConnectedSpace