# Is this set of equivalence classes a Tychonoff space?

I'm not sure why the space $(Y,\mathcal{V})$ described below is completely regular.

I start by taking $(S,\mathcal{U})$ and defining an equivalence relation $s\sim t\iff f(s)=f(t)$ for every continuous $f\colon S\to\mathbb{R}$. Let $Y$ be the set of such equivalence classes, and $\pi$ be the function mapping $s\in S$ to its equivalence class. So for continuous $f\colon S\to\mathbb{R}$, there is a unique $\phi(f)\colon Y\to\mathbb{R}$ such that $\phi(f)(\pi(s))=f(s)$. Then equip $Y$ with the weakest topology $\mathcal{V}$ such that each $\phi(f)$ is continuous, so every closed set in $Y$ has form $\bigcap_{i\in I}\phi(f_i)^{-1}(F_i)$ for some family $\{F_i\}$ of closed subsets of $\mathbb{R}$ and $\{f_i\}$ continuous.

With Brian M. Scott's aid, I've seen $(Y,\mathcal{V})$ is Hausdorff. But why is $(Y,\mathcal{V})$ completely regular? I take any closed set $A$ in $Y$, so $$A=\bigcap_{i\in I}\phi(f_i)^{-1}(F_i)$$ for some index set $I$. I pick some point $b\not\in A$, so there is some $i\in I$ such that $b\not\in \phi(f_i)^{-1}(F_i)$, that is, $\phi(f_i)(b)\not\in F_i$. Based on that, why is there some continuous function separating $A$ and $b$, with $A$ mapping into $\{0\}$ and $b$ mapping to $1$?

Is it then an easy consequence that the map $\pi$ is continuous as well? If not, perhaps I'll post that as a separate question later. Thanks!

## 1 Answer

I’ll use the same notation as in my previous answer. In particular, $\mathscr{F}$ is a separating family of real-valued functions on $Y$, and $\mathscr{V}$ is the coarsest topology making them continuous. Let $A\subseteq Y$ be closed and $b\in Y\setminus A$. Then $U = Y\setminus A$ is an open nbhd of $b$, so there must be a finite family $\{f_1,\dots,f_n\}\subseteq \mathscr{F}$ and a family $\{W_1,\dots,W_n\}$ of open subsets of $\mathbb{R}$ such that $$b \in \bigcap_{k=1}^n f_k^{-1}[W_k] \subseteq U\;.\tag{1}$$ The intersection is a basic open set in $\mathscr{V}$.

For $k=1,\dots,n$ let $g_k:Y\to\mathbb{R}:y\mapsto f_k(y)-f_k(b)$, and let $W_k' = \{x-f_k(b):x\in W_k\}$. Then $g_k(b)=0$ and $g_k^{-1}[W_k']=f_k^{-1}[W_k]$, so we can rewrite ${(1)}$ as $$b\in \bigcap_{k=1}^n g_k^{-1}[W_k'] \subseteq U\;.$$

Now choose $\epsilon>0$ so that $(-\epsilon,\epsilon)\subseteq W_k'$ for $k=1,\dots,n$, and let $J=(-\epsilon,\epsilon)$. Then $$b \in \bigcap_{k=1}^n g_k^{-1}[J] \subseteq U\;,$$ where the new intersection is again a basic open set in $\mathscr{V}$; call it $B$.

Let $g=\sum_{k=1}^n g_k$; $g$ is a continuous real-valued function on $Y$. Suppose that $y\in B$; then $|g_k(y)| < \epsilon$ for $k=1,\dots,n$, so $$|g(y)| = \left|\sum_{k=1}^n g_k(y)\right|\le \sum_{k=1}^n |g_k(y)| < n\epsilon\;.$$ Thus, $g(b)=0$, and $|g(a)|\ge n\epsilon$ for every $a\in A$. Can you finish it off from there?

Added: Recall that $\epsilon$ just had to be small enough, so we may assume that $\epsilon \le 1/n$ and hence that $n\epsilon \le 1$. Thus, $g(b)=0$, and $g(a)\ge 1$ for each $a\in A$. Now let $h:Y\to\mathbb{R}:y\mapsto \min\{|g(y)|,1\}$; clearly $h(b)=0$ and $h(a)=1$ for each $a\in A$, so $h$ does the trick provided that it’s continuous. (I see now that you wanted it to go the other way, but that’s no problem: just replace $h$ by $1-h$.)

You can show directly that $h$ is continuous, but in the long run it would be more useful to prove two general facts:

If $f$ is a continuous real-valued function on some space $X$, then $|f|$ and $\min\{f,1_X\}$ are continuous, where $1_X(x)=1$ for each $x\in X$.

(If you do it directly with $h$, you’re actually just going to find yourself proving a special casess of these facts anyway.) Do you want to take a stab at this, or shall I finish it off completely?

• Thanks again Brian! I'll try to prove the general fact in your added section. If I don't get it within a day or so I'll ask further. – Ashley Lin Oct 19 '11 at 23:17
• By the way, do you have any comment for why $\pi$ is continuous? By these results, for any $f\in S\to\mathbb{R}$ there is some $g\colon Y\to\mathbb{R}$, such that $f=g\circ\pi$. Then $\pi^{-1}=f^{-1}\circ g$. Since $f$ is continuous, the conclusion would follow if $g$ maps open sets to open sets, which I'm not sure is always true. If this is not trivial, I can post this as a new question. – Ashley Lin Oct 19 '11 at 23:35
• @Ashley: Let $U\subseteq Y$ be open, let $V=\pi^{-1}[U]$, fix $x\in V$, and let $y=\pi(x)\in U$. There are $f_1,\dots,f_n\in\mathscr{F}$ and open $W_1,\dots,W_n\subseteq\mathbb{R}$ s.t. $y\in\bigcap_{k=1}^n f_k^{-1}[W_k]\subseteq U$. Let $g_k=f_k\circ\pi$; each $g_k$ is continuous (since $f_k=\phi(g_k)$ in your notation), so $\bigcap_{k=1}^n g_k^{-1}[W_k]$ is an open nbhd of $x$ contained in $V$. Since $x$ was arbitrary, $V$ is open, and $\pi$ is continuous. – Brian M. Scott Oct 20 '11 at 0:00