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Urysohn's lemma states that if $A$, $B$ are disjoint closed sets in a normal space $X$, then there exists a continuous function $f : X \to [0, 1]$ such that $\forall\ a \in A$, $f(a) = 0$ and $\forall\ b \in B$, $f(b) = 1$.

How to apply Urysohn's lemma to prove that every normal space is completely regular? Or in fact, every $T_4$ space is a Tychonoff space?

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If $X$ is $T_4$, then $\{x\}$ is closed for every $x\in X$. Thus, if $F$ is a closed set not containing $x$, $\{x\}$ and $F$ are disjoint closed sets, and by Uryson’s lemma there is a continuous function $f:X\to[0,1]$ such that $f(x)=0$ and $f(y)=1$ for all $y\in F$. And that’s exactly what it means for $X$ to be Tikhonov.

It’s not actually true that every normal space is completely regular: the Sierpiński space $S$, whose underlying set is $\{0,1\}$ and whose open sets are $\varnothing,\{1\}$, and $\{0,1\}$, is a normal space that is not completely regular. It’s vacuously normal, since it does not contain two disjoint, non-empty closed sets. It’s not completely regular, because no continuous function $f:S\to[0,1]$ separates the point $1$ from the closed set $\{0\}$. To see this, suppose that $f$ is continuous and that $f(0)=1$; then $(0,1]$ is an open nbhd of $f(0)$ in $[0,1]$, so $f^{-1}\big[(0,1]\big]$ is an open nbhd of $0$ in $S$. But the only open set in $S$ containing $0$ is $S$ itself, so $f(1)=1=f(0)$.

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Thank you very much sir! – ccc Dec 27 '12 at 12:21
@ccc: You’re welcome! – Brian M. Scott Dec 27 '12 at 14:30

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