# How to make this construction rigorous, in order to prove that a topological space being normal and $T_1$ implies it being completely regular?

Let's start with some definitions, because I suspect that the ones we use in class aren't 100% universally accepted: If $$(T, \tau)$$ is a topological space, then I say that

• $$(T, \tau)$$ is normal, if for any two closed sets $$Z_1$$, $$Z_2$$, there exist open sets $$U_1$$, $$U_2$$ such that $$U_1 \supseteq Z_1$$, $$U_2 \supseteq Z_2$$ and $$U_1 \cap U_2 = \emptyset$$

• $$(T, \tau)$$ is $$T_1$$, if for any two points $$a,b \in T$$, there exist open sets $$U_1$$, $$U_2$$ such that $$a \in U_1,b \notin U_1$$, and $$a \notin U_2, b \in U_2$$

• $$(T, \tau)$$ is completely regular, if for any point $$t$$ and any closed set $$Z$$, there exists a continuous function $$f: T \to [0,1]$$ with $$f(t) = 1$$ and $$f(z) = 0$$ $$\forall z \in Z$$

• A dyadic number is a number of the form $$\frac{k}{2^n}$$, where $$k, n \in \mathbb N$$ (including zero) and $$k \leq 2^n$$. The set of dyadic numbers I will write as $$D$$. Obviously $$D \subseteq \mathbb Q \cap [0,1]$$.

A lemma we proved in class that I intend to use is:

• $$(T, \tau)$$ is $$T_1$$ $$\iff$$ every singleton set is closed.

And here's what I'm supposed to prove:

Proposition: If $$(T, \tau)$$ is normal and $$T_1$$ $$\implies$$ $$(T, \tau)$$ is completely regular

Idea: I remembered the construction of the devil's staircase function ($$s$$): there, it was sufficient to find the values of the input space where $$s = \frac 12$$, $$s = \frac 14$$, $$s = \frac 34$$ etc using the Cantor set's construction. Essentially, we only gave a set for every dayadic number, and the resulting function became continuous.

Proof (concept): Let $$(T, \tau)$$ be a normal, $$T_1$$ topological space, $$t \in T$$ be some point, and $$Z \subseteq T$$ is some closed subset. Obviously, $$f(t) = 1$$ and $$f(Z) = 0$$ must hold. Now, from the lemma, $$\{t\}$$ is closed, therefore we can use the normality property of $$T$$, which finds us two open subsets $$U_1 \supseteq \{t\}$$ and $$U_2 \supseteq Z$$ that are disjoint. We will set $$f(T \setminus (U_1 \cup U_2) ) = \frac 12$$. Visually, one cound say that we found a closed set that "separates" $$\{t\}$$ from $$Z$$. We will also say in advance that $$U_1$$ is where $$f > \frac 12$$ and $$U_2$$ is where $$f < \frac 12$$ holds (this will make the notation less clumsy in the following section)

We can repeat this process: Take the sets $$[f \geq \frac 12]$$ and $$Z$$ (which you could call $$[f = 0]$$ in a similar notation), apply normality, now we have two open sets that will be the sets where $$f > \frac 14$$ and $$f < \frac 14$$ respectively. The set $$T \setminus ([f > \frac 14] \cup [f < \frac 14])$$ is now a closed set yet again, so we will be able to continue this iteration infinitely many times.

Questions:

(1) Is this function now well-defined? I mean, after this process, I can define $$f$$ on any point $$x \in T$$: $$f: x \mapsto \inf \{ d \in D | x \in [f > d] \}$$... Or $$f: x \mapsto \sup \{ d \in D | x \in [f < d] \}$$? Which definition would be more sensible? Is there a difference? Are they equal?

(2): Why is this function continuous? I 'feel' that it should be, and the construction doesn't seem to show any 'tearing', but I'm not able to formally prove that it is so, not with the "preimage of every open set is open" notion of continuity or with the "continuity at each point" notion.

• It's immediate from Urysohn's lemma for normal spaces. Didn't you do that in class? Commented Mar 12, 2021 at 8:31
• No, this statement is actually included in our version as well. It seems all I really did was reinvent the wheel here :( Commented Mar 12, 2021 at 9:16
• @maritsm ... the professor call this exercise "an especially difficult one, only for the most ambitious" ... Don't you think it is subtle irony? Commented Mar 12, 2021 at 9:51
• I don't think it was irony, as the exercise was presented before the Urysohn lemma was. I think the intention might have been to think about this before we covered Urysohn Commented Mar 12, 2021 at 9:53
• @maritsm: That was my suspicion when I saw the question: that it was preparation for the Urysohn lemma. Commented Mar 12, 2021 at 19:42