Every subset of a compact Hausdorff space is completely regular

Question

I'm trying to convince myself on this statement.

Suppose $X$ is compact and Hausdorff, and let $U\subseteq X$ be any subset. Let $x\in U$ and let $C\subset U$ be closed in the subspace topology with $x\notin C$. Then $C$ is also closed in $X$.

How am I suppose to define a continuous function $f:U\rightarrow [0,1]$ such that $f(x)=0$ and $f(y)=1$ for each $y\in C$?

score 2 · Accepted Answer · answered Oct 15 '17 at 11:14

2

Since X is compact Hausdorff, it is normal
thus completely regular. Finally notice that
every subspace of a completely regular space
is completely regular.

answered Oct 15 '17 at 11:14

William Elliot

18,025

But why does compact and Hausdorff imply normal? – Sid Caroline Oct 15 '17 at 11:18
Compact Hausdorff implies regular. Compact Hausdorff regular implies normal. – Angina Seng Oct 15 '17 at 11:19
1

@SidCaroline see answers to this – Henno Brandsma Oct 15 '17 at 11:37

Every subset of a compact Hausdorff space is completely regular

1 Answers1