A topological space $X,\tau$ is said to be normal space if for each pair of disjoint closed sets $A$ and $B$, there exists open sets $U$ and $V$ such that $A\subseteq U$,$B\subseteq V$ and $U\cap V=\emptyset$. Prove that every metrizable space is normal space.
If $X,\tau$ is a metrizable space then there exists $d:X\times X\to[0,+\infty]$ that defines the open sets in $\tau$.
Consider $\epsilon=\frac{d(a,b)}{2}\forall a\in A$ and $b\in B$.
Then $A\subset \bigcup_{a\in A} \mathscr{B}(a,\epsilon)$ since A is closed. The affirmation is proven in the following way: some $a\in Fr(A)$ then $B(a,\epsilon)\cap Ext(A)\neq\emptyset$
In the same way $B\subset \bigcup_{b\in B} \mathscr{B}(b,\epsilon)$
If $U=\bigcup_{a\in A} \mathscr{B}(a,\epsilon)\\V=\bigcup_{b\in B} \mathscr{B}(b,\epsilon)$,
then $U\cap V=\emptyset$.
Therefore $(X,d)$ is a normal space.
Question:
Is my proof right? If not. Why?
Thanks in advance!
Best Answer
The proof is wrong, as "$\varepsilon = \frac{d(a,b)}{2}\forall a \in a, \forall b \in B$" is ill-defined (the infimum of those numbers can be $0$ for disjoint closed sets).
Another, better idea is to note that $f_A: x \to d(x,A)$ is a continuous function from $X$ to $\mathbb{R}$ (where $d(x,A) = \inf \{ d(x,a): a \in A\}$ and which obeys $d(x,A) = 0$ iff $x \in \overline{A}$.
Then for disjoint closed $A$ and $B$, the function $$f(x) = \frac{d(x,A)}{d(x,A) + d(x,B)}$$ is well-defined (as no point $x$ exists such that $d(x,A) + d(x,B) = 0$, so that this term is always $>0$) and also continuous. Moreover $f[A] = \{0\}$ and $f[B]= \{1\}$ so that $U=f^{-1}[(-\infty,\frac13)]$ and $V = f^{-1}[(\frac{2}{3}, +\infty)]$ are disjoint open subsets of $A$ resp. $B$.
And $$A=(f_A)^{-1}[\{0\}] = \bigcap_n f^{-1}[(-\infty,\frac1n)]$$ shows that all closed sets are $G_\delta$'s and $(X,d)$ is even perfectly normal.