Proving every metrizable space is normal space

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?

• Looks good to me. – Yanko Nov 10 '18 at 15:37

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.

• I am sorry but I am not understanding your proof. – Pedro Gomes Nov 10 '18 at 16:12
• Could you explain what is the problem of the infimum of metric in my proof being 0? – Pedro Gomes Nov 10 '18 at 16:13
• @PedroGomes you cannot have a ball of radius $0$. – Henno Brandsma Nov 10 '18 at 16:14
• @PedroGomes what don't you understand? There are lots of posts showing the continuity of $d(x,A)$ as a function of $x$. – Henno Brandsma Nov 10 '18 at 16:15
• I have proven that continuity. But I do not understand why you define the metric $f(x) = \frac{d(x,A)}{d(x,A) + d(x,B)}$ and the reson why f(A)=0 and f(B)=1. And I cannot understand the link between the conclusion and the original goal of proving the space is normal. – Pedro Gomes Nov 10 '18 at 16:19

To give another proof more along the lines you started out with, using unions of open balls centered at the points of the sets:

Suppose we have $$A,B$$ disjoint and closed in a metric space $$(X, d)$$.

For each $$a \in A$$ we have that we have $$r_a>0$$ such that the open ball $$B(a, r_a)$$ is disjoint from $$B$$. This follows from $$B$$ being closed and $$a \notin B$$. So for these $$r_A$$ we have the property $$\forall a \in A : \forall b \in B: d(a,b) \ge r_a\tag{a}$$

Likewise for each $$b \in B$$ we have some $$s_b>0$$ such that $$B(b,s_b) \cap A=\emptyset$$ or

$$\forall b \in B: \forall a\in A: d(b,a) \ge s_b\tag{b}$$

Now define $$O_A=\bigcup\{B(a,\frac{r_a}{2}): a \in A\}$$ and $$O_B=\bigcup\{B(b, \frac{s_b}{2}): b \in B\}$$, which are both open as unions of (basic) open balls. Clearly $$A \subseteq O_A$$ and $$B \subseteq O_B$$.

Suppose that $$p \in O_A \cap O_B$$, which means that for some $$a_0 \in A$$ and some $$b_0 \in B$$ we have that $$p \in B(a_0, \frac{r_{a_0}}{2})$$ and $$p \in B(b_0, \frac{s_{b_0}}{2})$$. Define $$r=\max(r_{a_0}, s_{b_0})$$, and by $$(a)$$ and $$(b)$$ we have that $$r \le d(a_0, b_0)$$,

But then: $$r \le d(a_0, b_0) \le d(a_0,p)+ d(p, b_0) < \frac{r_{a_0}}{2} + \frac{s_{b_0}}{2} \le \frac{r}{2} + \frac{r}{2}=r$$

We now have the contradiction $$r < r$$ so the intersection of $$O_A$$ and $$O_B$$ is empty and we have shown normality.