For each Hausdorff space, a completely regular quotient space. Problem from Dugundji's Topology [mistake on the problem].

Question

I got stuck on problem 1, section 7, Chapter VII (Separation Axioms) of Dugundji's Topology.

1. Let $Y$ be Hausdorff, and let $R$ be the relation: $x\sim y$ if there is no continuous $\varphi\colon Y\to I$ with $\varphi(x)\neq \varphi(y)$.
   a. Show that $R$ is an equivalence relation in $Y$.
   b. Show that $Y/R$ is completely regular.

So we have, where $I^Y$ are the continuous functions from $Y$ to $I:=[0,1]$, $$R=\big\{(a,b)\subseteq Y^2\mid \{\varphi\in I^Y\mid \varphi(a)\neq\varphi(b)\}=\emptyset\big\}.$$ a. Is trivial.
b. My attempt: Let $[p]\in Y/R$ and $A\subseteq Y/R$ closed with $[p]\notin A$. This implies $\forall [q]\in A,\; q\nsim p$. For each $[q]\in A$, let $\varphi_{[q]}\in I^Y$ be such that $\varphi_{[q]}(p)\neq \varphi_{[q]}(q)$ (WLOG, $1=\varphi_{[q]}(p)\neq \varphi_{[q]}(q)=0$). Note that, $\forall [q]\in A$, if $x\sim y$, then $\varphi_{[q]}(x)=\varphi_{[q]}(y)$ (since $\varphi_{[q]}\in I^Y$). That is, $\varphi_{[q]}\circ p^{-1}$ is single valued (where $p\colon Y\to Y/R$ is defined by $y\mapsto [y]$ is the identification map) and therefore continuous.

Let $\phi_{[q]}=\varphi_{[q]}\circ p^{-1}$, then $$\begin{align}\inf_{[q]\in A}\phi_{[q]}\colon Y/R&\to I\\ [y]&\mapsto \inf\{\varphi_{[q]}(y)\mid [q]\in A\}\end{align}$$ is an upper semi continuous map with $\inf_{[q]\in A}\phi_{[q]}([p])=1$ and $\forall [q']\in A,\; \inf_{[q]\in A}\phi_{[q]}([q'])=0$. This is as close as I could get (note that I didn't use the fact that $A\subseteq Y/R$ was closed and that Y was Hausdorff).

The question: I'm looking for any hints (or solutions at this point) to the problem. It's not necessary to continue or do something related to my attempt, I'm stuck and looking for any way of prooving it.

Answer 1

As David Hartley points out is his comment, thanks by the way, if $Y$ is completely Hausdorff, then $Y\cong Y/R$. Being the case that there exist completely Hausdorff spaces which are not completely regular, the theorem suggested by the problem is false.

In my edition of the book (1966), problem 1 of section 7 on chapter VII (p. 159) corresponds to the one shown in the my question. In later versions (such as the twelfth printing, 1978) another problem is proposed instead:

1. Let $Y$ be a $T_0$ space satisfying the condition: each $y\in Y$ and closed $A$ not containing $y$ have disjoint nbds. Prove: $Y$ is a regular (Hausdorff) space.

This suggests that a mistake was made in older copies of the book: Dugundji be trolling.

It's easy to adapt my attempted proof to show that $Y/R$ is always completely Hausdorff, $\phi_{[q]}$ being the separating map.