Completely Regular Reflection

I have tried to construct the completely regular reflection of any Topological space. I am not sure whether my argument is precise or I made a mistake somewhere. Especially for the choice of $$(X,T_{T})$$ and uniqueness of $$h$$.

Definition: A topological space $$X$$ is said to be completely regular if for each $$x\in X$$ and closed set $$F$$ not containing $$x$$, there is a continuous function $$f:X\longrightarrow [0,1]$$ such that $$f(x)=1$$ and $$f(F)=\{0\}$$.

This is how I have constructed it: Let $$(X,T)$$ be a topological space. Let $$T_{T}$$ be a topology on $$X$$ whose basic closed sets are zero-sets of $$(X,T)$$. Consider $$id_{X}:(X,T)\longrightarrow (X,T_{T})$$. Claim: $$(X,T_{T})$$ is the completely regular reflection of $$X$$ and $$id_{X}$$ the reflection map.

$$(X,T_{T})$$ is Tychonoff: This follows from the fact that zero-sets of $$(X,T)$$ form a base for the closed sets on $$(X,T_{T})$$.

Continuity: Let $$A$$ be a closed set of $$(X,T_{T})$$. Then $$A=\bigcap[f^{-1}(\{0\}):f\in C(X)]$$. But each zero-set is closed, so $$A$$ is closed in $$(X,T)$$.

Let $$f:(X,T)\longrightarrow (X',T')$$ be a continuous function from $$(X,T)$$ to a compleltely regular space $$(X',T')$$.
Define $$h:(X,T_{T})\longrightarrow (X',T')$$ by $$x\mapsto f(x)$$. It is clear that $$h$$ is well-defined. $$h$$ is continuous: Let $$A$$ be closed subset of $$X'$$. Then $$A=\bigcap[g^{-1}(\{0\}):g\in C(Y)]$$. zero-sets are preserved by continuous pre-images, so $$f^{-1}(A)=\bigcap[(g\circ f)^{-1}(\{0\}):g\circ f\in C(X)]$$ which implies that $$f^{-1}(A)$$ is closed in $$(X,T_{T})$$. But $$f^{-1}(A)=h^{-1}(A)$$. Thus $$h$$ is continuous.

Observe that $$h(id_{X}(x))=f(x)$$.

$$h$$ is unique: Let $$k$$ be another continuous map from $$(X,T_{T})$$ to $$(X',T')$$ satisfying $$k\circ id_{X}=f$$. Then for each $$x\in X$$, $$k(x)=f(x)=h(x)$$. #####

• It works if Tychonoff assumes no $T_1$, but just functional separation (what I would prefer to call complete regularity). – Henno Brandsma Nov 9 '18 at 16:10
• Thank you @Henno. I must edit it and write completely regular. – Percy Nov 9 '18 at 19:04
• "Must edit" is an overstatement. There are also authors who include $T_1$ in the definition of "completely regular". – Andreas Blass Nov 10 '18 at 1:07

For me, Tychonoff ($$T_{3\frac12}$$) is $$T_1$$ plus completely regular. $$X$$ is completely regular when for all closed sets $$C\subseteq X$$ and all $$x \in X\setminus C$$ we have a continuous $$f: X \to \mathbb{R}$$ such that $$f(x) = 0$$ and $$f[C] = \{1\}$$. The reflection of taking the zero-sets as a closed base works for complete regularity. There are even $$T_3$$ spaces $$X$$ for which the set of zero-sets is just the indiscrete topology, because all real-valued functions are constant on it. So we don't always get a very nice space.
That the cozero-set topology $$\mathcal{T}_{\textrm{coz}}$$ is completely regular follows because all real-valued functions that were continuous on $$X$$ are still continuous for $$(X,\mathcal{T}_{\textrm{coz}}$$).