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 $x\in X$, $k(x)=f(x)=h(x)$. #####

  • $\begingroup$ It works if Tychonoff assumes no $T_1$, but just functional separation (what I would prefer to call complete regularity). $\endgroup$ – Henno Brandsma Nov 9 '18 at 16:10
  • $\begingroup$ Thank you @Henno. I must edit it and write completely regular. $\endgroup$ – Percy Nov 9 '18 at 19:04
  • $\begingroup$ "Must edit" is an overstatement. There are also authors who include $T_1$ in the definition of "completely regular". $\endgroup$ – 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 $(X,\mathcal{T}_{\textrm{coz}}$).


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