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Prove the following result without using Urysohn's lemma.

$(X,\mathscr T)$ is normal and every closed subset of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal.

My effort: I have proved using Urysohn's lemma. How do I prove without the use of Urysohn's lemma?

Let $A$ and $B$ be disjoint closed sets of $X$. My aim is to construct a continuous function from $X\to [0,1]:f^{-1}{(0)}=A$ and $f^{-1}{(1)}=B.$ Using Urysoh's lemma resultfollows trivially.

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    $\begingroup$ The definition of perfect normality is $f^{-1}(0)=A$ and $f^{-1}(1)=B$, which is stronger than $f|_A≡0$ and $f|_B≡1$ $\endgroup$ – YuiTo Cheng Mar 20 at 2:07
  • $\begingroup$ sorry!I will correct it $\endgroup$ – Math geek Mar 20 at 3:11
  • $\begingroup$ The proof of Urysohn's Lemma can be easily modified to obtain your theorem. $\endgroup$ – YuiTo Cheng Mar 20 at 4:19
  • $\begingroup$ Can I have shorter proof independent of Urysohn's lemma? $\endgroup$ – Unknown x Mar 20 at 11:06
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This question is very similar, maybe you're using the same text?

As I said there: you really cannot avoid Urysohn's lemma there, or you'd have to reprove it, with extra modifications. What other ways do you have to construct functions on a normal space?

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  • $\begingroup$ But my teacher asked to prove independent of Urysohn's lemma. $\endgroup$ – Math geek Mar 21 at 1:12
  • $\begingroup$ @Mathgeek then redo its proof with modifcations.. $\endgroup$ – Henno Brandsma Mar 21 at 4:58
  • $\begingroup$ you mean there is no alternative! $\endgroup$ – Math geek Mar 21 at 15:02
  • $\begingroup$ @Mathgeek No, I don't see an alternative. Nor has anyone else so far. $\endgroup$ – Henno Brandsma Mar 21 at 16:23
  • $\begingroup$ @Mathgeek If this lemma holds, I think you can rather easily reprove Urysohn's lemma from it. $\endgroup$ – Henno Brandsma Mar 21 at 16:30
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First, I think that the task “prove $X$ without using $Y$” can mean two different things:

  1. You are actually supposed to reprove $Y$ or a special case of $Y$ to obtain $X$. I can remember an exam where we were supposed to prove something that was in fact a special case of the Banach fixed-point theorem. Of course we weren't allowed to use the Banach fixed-point theorem as a black box. The actual task was to reprove it in the special case.
  2. You really want to find a conceptually different proof of $X$. And here I completely agree with Henno Brandsma that there may be no alternative… especially when you seek for shorter and/or simpler proof. I give some reasons in the comments below.

First, let us fix sume definitions. A topological space $X$ is

  • normal if every disjoint closed sets $A, B ⊆ X$ can be separated by disjoint neighborhoods;
  • functionally normal if every disjoint closed sets $A, B ⊆ X$ can be functionally separated, i.e. there is a continuous function $f\colon X \to [0, 1]$ such that $A ⊆ f^{-1}(0)$ and $B ⊆ f^{-1}(1)$;
  • perfectly normal if every disjoint closed sets $A, B ⊆ X$ can be perfectly functionally separated, i.e. there is a continuous function $f\colon X \to [0, 1]$ such that $A = f^{-1}(0)$ and $B = f^{-1}(1)$.
  • ultranormal if every disjoint closed sets $A, B ⊆ X$ can be separated by a clopen set, i.e. there is a clopen set $C ⊆ X$ such that $A ⊆ C$ and $C ∩ B = ∅$ – equivalently, $X$ has large inductive dimension zero.

Clearly, perfect normality implies functional normality, and functional normality implies normality. The fact that normality also implies functional normality is the Urysohn lemma.

We want to prove that a normal space such that every closed set is $G_δ$ is perfectly normal. I don't agree with the OP that it follows trivially from Urysohn's lemma. We still need to prove that a closed $G_δ$ set in a functionally normal space is functionally closed, which boils down to the fact that functionally closed sets are stable under countable intersections – easy, but not trivial.

As said before, I don't believe there is a simple proof not using the ideas of the proof of the Urysohn's lemma. We need to construct a continuous function that may attain countiuum many values, so we probably need to use the normality to get a “densely linearly ordered family of separations”, which is already the core of the proof of Urysohn's lemma.

Another reason, as Henno Brandsma suggests, is that the Urysohn's lemma follows from the theorem if stated as “every closed $G_δ$ subset of a normal space is functionally closed”. This is because in a normal space $X$ for every closed set $F ⊆ X$ and open set $U ⊆ X$ such that $F ⊆ U$ there is a closed $G_δ$ set $G ⊆ X$ such that $F ⊆ G ⊆ U$.

What we can do is to prove the theorem directly by using the construction in the proof of Urysohn's lemma and taking an densely linearly ordered family of separations such that $A$ is its intersection, so it will be precisely the zero set of the resulting function.

Also note that proving the theorem for an ultranormal space is indeed easier. Of course an ultranormal space is functionally normal, and so the theorem for ultranormal spaces follows as before without using the Urysohn's lemma. But we may also prove it directly – every countable intersection of clopen sets is functionally closed, and it is easy to explicitly write down the function.

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