Normal Hausdorff implies Tychonoff

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) uses::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.