Is $\mathbb R$ completely normal space?

Question

Is $\mathbb R$ completely normal ( $T_5$ ) space?

I've seen this Wiki Page about Urysohn's lemma, which states that a topological space is normal if and only if any two disjoint closed sets can be separated by a continuous function.

Two plain subsets $A$ and $B$ are said to be separated by a function if there exists a continuous function $f : X \to[0,1]$ such that $f(A) = 0$ and $f(B) = 1$. Any such function is called a Urysohn function for $A$ and $B$.

So, now I need to construct continuous function $f : \mathbb R \to [0, 1]$ such that for all closed $A, B \subseteq \mathbb R: A \cap B = \varnothing,$ we have $f(A)=0$ and $f(B)=1$.

However, I wonder if one could construct such continuous function. I tried some thoughts by brute force but no success. Is there any algorithm to solve this kind of problems?

Thanks in advance!

Answer 1

Every metric space is normal. Since every subspace of a metric space is a metric space, it thus follows that every metric space is completely normal.

You seem to be asking for a proof that every metric space is perfectly normal. Let's just go ahead and write something down.

Fix a metric space $(X,d)$. If $A\subset X$ is closed, then $$f_A:x\mapsto d(x,A)=\inf\{d(x,a)\mid a\in A\}$$ is a continuous function $X\rightarrow[0,\infty)$ with $f^{-1}_A(0)=A$. To see that this function is continuous just use the triangle law to show that it is even Lipschitz.

Suppose $B\subseteq X$ is closed and disjoint from $A$. Form $f_B$ as above. Now consider $$h=\frac{f_A}{f_A+f_B}.$$ Since $A\cap B =\emptyset$ the function is well-defined. Clearly $h^{-1}(0)=A$ and $h^{-1}(1)=B$.

(Edit: To put my open comment in contex note that a space is completely normal if and only if it is hereditarily normal. This is a standard definition of completely normality. The equivalence of the two conditions is stated on the wikipedia page. I've shown that each metric space is perfectly normal. Every perfectly normal space is completely normal. For istance with the notation above consider the disjoint neighbourhoods $U_A=h^{-1}[0,1/3)$ and $U_B=h^{-1}(2/3,1]$.)

Answer 2

In fact, $\Bbb{R}$ is not only $T_5$, it's $T_6$.

If you surf from the Urysohn's lemma Wikipedia page over to the Wikipedia page for "Normal space," you learn:

All metric spaces (and hence all metrizable spaces) are perfectly normal Hausdorff

Since $\Bbb{R}$ is a metric space, and being perfectly normal is stronger than being completely normal, $\Bbb{R}$ is completely normal.