Normal Hausdorff implies Tychonoff

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., normal Hausdorff space) must also satisfy the second topological space property (i.e., Tychonoff space)
View all topological space property implications | View all topological space property non-implications
Get more facts about normal Hausdorff space|Get more facts about Tychonoff space

Statement

Any normal space is a Tychonoff space, i.e., it is both a completely regular space and a Hausdorff space.

Related facts

• Locally compact Hausdorff implies completely regular

Converse

• Completely regular not implies normal

Facts used

1. Urysohn's lemma: This states that if $X$ is a normal space and $A$ and $B$ are pairwise disjoint closed subsets, there is a continuous function $f:X\to [0,1]$ such that $f$ takes the value $0$ everywhere on $A$ and $f$ takes the value $1$ everywhere on $B$.

Proof

Given: A normal space $X$, a point $p\in X$ and a closed subset $A\subseteq X$ such that $p$ is not contained in $A$.

To prove: $X$ is $T_{1}$, and there is a continuous function $f:X\to [0,1]$ such that $f(p)=1$ and $f(a)=0$ for all $a\in A$.

Proof: By definition, normal spaces are $T_{1}$, so $X$ is $T_{1}$: its points are closed. Hence, we can take $A=A$ and $B=\{p\}$ and apply Urysohn's lemma, to obtain the required continuous function.