I have some trouble finding a right way to solve this problem. The only thing I can think of is the Tietze Extension Theorem, but somehow it seems that the converse of Tietze Extension Theorem is more useful.

Suppose $X$ is Tychonoff $(T_1)$ and $X$ has the property that every continuous, bounded real valued function on a closed subset of $X$ has a continuous extension to all of $X$. We want to show that $X$ is normal.

My attempt: Since $X$ is Tychonoff, then for any $x,y\in X$ with $x\neq y$, there is a neighbourhood $U$ of $x$ such that $x\in U$ but $y\notin U$. Let $f$ be any continuous, bounded real valued function on a closed subset $Y$ of $X$, i.e. $f:Y\to \mathbb{R}$ and its continuous extension be $F:X\to\mathbb{R}$.

To show $X$ is normal, let $A$ be a closed subset of $X$ and $X\cap A=\emptyset$. I am not sure how can we continue from here.

How can we construct the neighbourhoods of $X$ and $A$? How can we make use of $f$ and $F$?

Could anyone please give some concrete hints? I have spent some time on this problem but still couldn't see any light. Thanks.

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    $\begingroup$ One version of Tietze's theorem is that a space is $T_4$ if and only if every bounded real-valued function defined on a closed subset can be continuously extended to the whole space. That the extension property implies $T_4$ is the easy direction. $\endgroup$ – Daniel Fischer Sep 28 '16 at 12:08
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    $\begingroup$ @DanielFischer Thanks, but I have never seen that Theorem before. Is there any other way besides using that version of Tietze's theorem? $\endgroup$ – user338393 Sep 28 '16 at 12:13
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    $\begingroup$ As I said, what you need is the easy direction, that the extension property implies $T_4$ [I'm assuming that you use the terminology where "normal = $T_4 + T_1$"]. Take two disjoint closed subsets $A,B$ of $X$. Which $f \colon A \cup B \to [0,1]$ may help getting you a separation of $A$ and $B$ via extension? $\endgroup$ – Daniel Fischer Sep 28 '16 at 12:17

Let $A,B$ be two disjoint closed subsets of $X$.

check:$f:A\cup B\rightarrow \mathbb{R} $ with $f(A)=\{0\}$ and $f(B)=\{1\}$ is a continuous function.

So by assumption, you get the continuous extension $F:X\rightarrow \mathbb{R} $ with $F|_{A\cup B}=f$, then consider $F^{-1}((-\frac{1}{2},\frac{1}{2}))$ and $F^{-1}((\frac{1}{2},2))$, which are disjoint open sets separating $A$ and $B$.

That means $X$ is normal.

  • $\begingroup$ I am wondering where do we use the $T_1$ assumption on $X$, or do we need it? $\endgroup$ – awllower Nov 6 '16 at 16:56
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    $\begingroup$ No!~We don't need it! $\endgroup$ – yoyo Nov 7 '16 at 1:08

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