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I am doing exercise 33.5 of Munkres:

Exercise 33.6: Theorem (Strong form of Urysohn Lemma). Let $X$ be a normal space. There is a continuous function $f : X \to [0,1]$ such that $f(x) = 0$ for all $x\in A$ , and $f(x) = 1$ for all $x \in B$, and $0 < f(x) < 1$ otherwise iff $A$ and $B$ are disjoint closed $G_\delta$ sets in $X$.

Now the existence of such a function immediately gives that $A$ and $B$ are closed $G_\delta$ sets since $$A = f^{-1}\left( \bigcap_{n\geq 1} B_{\frac{1}{n}}(0)\right)$$

and similarly for $B$. Now for the other direction, I have also proved it but only using the fact that one of $A$ or $B$ is $G_\delta$. How did I do this? Well by exercise 4 of the same section there is $f : X \to [0,1]$ such that $f(x) = 0$ for all $x \in A$ and $f(x) > 0$ for all $x \notin A$ since $A$ is $G_\delta$. By the Urysohn Lemma, there is $g(x) : X\to [0,1]$ such that $g|_B = 0$ and $g|_A = 1$. Then $$h(x) = \frac{g(x)}{f(x) + g(x)}$$ is the required function.

My question is: For this direction of the problem, is it a superfluous condition that we need both $A$ and $B$ to be $G_\delta$?

Thanks.

1 Answers1

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No, though for many common spaces it is a superfluous condition.

A subset $A$ of a topological space is called a zero-set if there is a continuous function $f : X \to \mathbb{R}$ such that $A = f^{-1} [ \{ 0 \} ]$. From Urysohn's Lemma one can almost immediately conclude that in a normal space the zero-sets are exactly the closed G$_\delta$ sets.

Added: Note that in the statement of the Strong Form of Urysohn's Lemma we require $f : X \to [ 0 , 1 ]$ to be continuous, $A = f^{-1} [ \{ 0 \} ]$ and $B = f^{-1} [ \{ 1 \} ]$. This means that both $A$ and $B$ must be zero-sets, and so by the statement above they must be closed G$_\delta$.

There is a class of spaces strictly smaller than the normal spaces called perfectly normal, defined by being normal and having all closed sets G$_\delta$. In these spaces the G$_\delta$ condition is clearly superfluous, and many of the most common normal spaces are actually perfectly normal: the real line (in fact all metric spaces), and the Sorgenfrey line, for example.

As an example of a normal not perfectly normal space, consider the one-point compactification of an uncountable discrete space: $\alpha X = X \cup \{ \infty \}$ where $X$ is uncountable, and topologised so that all points of $X$ are isolated, and the open neighbourhoods of $\infty$ are of the form $\{ \infty \} \cup A$ where $A$ is a co-finite subset of $X$.

A basic fact about this space is that given any continuous function $f : \alpha X \to \mathbb{R}$, the set $$\{ x \in X : f(x) \neq f ( \infty ) \}$$ is countable. It follows that the closed set $\{ \infty \}$ is not a zero-set.

Added: In particular, given any finite (nonempty) $F \subseteq X$ there is no continuous function $f : \alpha X \to [ 0 , 1 ]$ such that $f^{-1} [ \{ 0 \} ] = \{ \infty \}$ and $f^{-1} [ \{ 1 \} ] = F$.

user642796
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  • Dear Arthur, in my proof above I did not use the fact that $B$ was $G_\delta$. Is my proof wrong then? –  Mar 25 '13 at 12:22
  • @伟轩: There is nothing in your proof to guarantee that $f(x) < 1$ for all $x \notin B$. As long as $g(x) = 0$ you will have $h(x) = 1$, and we only know that $B \subseteq g^{-1} [ { 1 } ]$. (I think there is a typo in the definition of $h$ in the OP.) – user642796 Mar 25 '13 at 12:29
  • @伟轩: Okay, I see that you defined $h$ in the opposite manner that I thought, but my objection remains, only reversed. Your definition of $h$ clearly has that $A = h^{-1} [ { 1 } ]$, and $h^{-1} [ { 0 } ] = g^{-1} [ { 0 } ] \supseteq B$. But, again, the set $g^{-1} [ { 0 } ] \setminus B$ could be nonempty. – user642796 Mar 25 '13 at 12:39
  • Right. But I still have a function $ j : X \to [0,1]$ such that $j(x) = 0$ for all $ x \in B$, and positive otherwise. Perhaps I have to modify my $h$. –  Mar 25 '13 at 12:50
  • @伟轩: Yes, you will have to use the fact that $B$ is also G$_\delta$ somewhere in your proof. To be honest, the solution in the OP is 99.99999% correct. All that is missing is a different condition on the function $g$. – user642796 Mar 25 '13 at 12:55
  • Dear Arthur, you may address me as @BenjaLim if the chinese characters don't render. Now I think we know that there is $g : X \to [0,1]$ such that $g(x) = 0$ on $B$ and positive otherwise. I think $h(x) = \frac{g(x)}{g(x) + f(x)}$ now works. –  Mar 25 '13 at 13:00
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    @BenjaLim: Yup, exactly correct! – user642796 Mar 25 '13 at 13:19
  • Sometimes I do stupid things like what I asked in my answer above :) –  Mar 26 '13 at 09:13
  • @BenjaLim: No worries: we all do stupid things! (At least I know I do almost every day! :-) ) After looking at my answer I found it to be quite lacking. What I should have said, but didn't, is now part of the addenda. – user642796 Mar 26 '13 at 10:28