# Show $A \subseteq X$, $X$ normal, is separated by a continuous function iff $A$ is closed $G_\delta$

Verify the claim: Let $A \subseteq X$, $X$ is normal. Then $A$ is separated by a continuous function $$f:X \to [0,1], f(x) = 0, \forall x \in A, f(x) >0, \forall x \notin A$$ iff $A$ is closed, $G_\delta$ set

I think I will be required to use Urysohn Lemma somehow, so:

(Urysohn Lemma) $(X, \mathcal{T})$ is normal iff for all $C,D \subseteq X$, where $C,D$ are disjoint nonempty closed sets, there exists continuous function $$f:X \to [0,1], f(x) = 0, \forall x \in C, f(x) =1 , \forall x \in D$$

Proof attempt:

$(\Rightarrow)$

• Given $A \subseteq X$
• First show it is closed
• Suppose there exists $$f:X \to [0,1], f(x) = 0, \forall x \in A, f(x) >0, \forall x \notin A$$
• Since $\{0\}$ is a closed singleton, and $f$ is continuous, therefore $f^{-1}(0) = A$, is closed.
• Now show it is $G_\delta$
• (Normality) Since $X$ is normal, $A \subset X$, there exists $U_1 \in \mathcal{T}$ such that $A \subset U_1 \subset \overline U_1 \subset X$
• Repeated application of normality, we produce countable set of open sets $\{U_n|n \in \mathbb{N}\}$
• Hence $A = \bigcap\limits_{n \in \mathbb{N}} U_n$, so it is $G_\delta$

$(\Leftarrow)$ I find this direction not so easy

• Given $A \subseteq X$, where $A$ is closed and $G_\delta$
• Let $C \subset X\backslash A$ be a closed set.
• By Urysohn Lemma, for all $A,C \subseteq X$, where $A,C$ are disjoint nonempty closed sets, there exists continuous function $$f:X \to [0,1], f(x) = 0, \forall x \in A, f(x) =1 , \forall x \in C$$
• However $X\backslash A$ is not closed...otherwise we are done.
• ???
• But we know that by normality of $X$, there exists disjoint open sets $U,V$ such that $A \subset U$, $C \subset V$
• Since $A = \bigcap\limits_{n \in \mathbb{N}} U_n$, take $\{V_n|n \in \mathbb{N}\}$ be open sets containing $C$ such that $V_n$ is disjoint from $U_n$ $$\vdots$$

Can someone verify my top proof and provide some suggestions as to how I should continue with the reverse direction. Thank you!

• ???$\Rightarrow$Profit? – Luiz Cordeiro Aug 8 '16 at 2:28

Your proof that $A$ is closed is fine. I think you're being overly complicated with your proof that $A\in G_\delta$. Simply note that:

$$A=\bigcap_{n=1}^\infty f^{-1}([0,1/n))$$

For the converse, suppose $A=\bigcap_n U_n$, with $U_n$ open. Fix $n$. Then $A\subset U_n$, so $A$ and $U_n$ are disjoint open sets. By normality, we can apply Urysohn's lemma to infer the existence of a continuous $f_n:X\to [0,1]$ such that $f_n(A)=\{0\}$ and $f_n(U_n^c)=\{1\}$.

Apply this to each $n$, and put:

$$f=\sum_{n=1}^\infty 2^{-n} f_n$$

I leave to you the verification that this $f$ works. Hover your cursor below to get a further hint:

Since each $f_n$ is bounded by $1$, this series converges uniformly, and by the Weierstrass M-test, $f$ is continuous.