2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ No!~We don't need it! $\endgroup$ – yoyo Nov 7 '16 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.