Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to verify some topological properties to see that for any topological space $(S,\mathcal{U})$, there is a continuous surjection $\pi\colon S\to T$ for $(T,\mathcal{V})$ a Tychonoff space, such that any continuous real valued function $f$ on $S$ can be expressed as $g\circ\pi$ for continuous real valued $g$ on $T$ and $\pi\colon S\to T$ continuous and surjective.

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.

My question is, why is $(Y,\mathcal{V})$ Hausdorff? I let $\pi(s)$ and $\pi(t)$ be distinct equivalence classes. Some observations I have are: $\phi(f)(\pi(s))=f(s)$ and $\phi(f)(\pi(t))=f(t)$. So $\pi(s)\in\phi(f)^{-1}(f(s))$ and in fact $\pi(s)\in\phi(f)^{-1}(f(z))$ for each $z\sim s$. By the same reasoning, $\pi(t)\in\phi(f)^{-1}(f(w))$ for each $w\sim t$. Also, $\phi(f)^{-1}(f(w))\cap\phi(f)^{-1}(f(z))=\emptyset$ since $f(z)\neq f(w)$.

This is as far as my reasoning takes me. What's a way to construct disjoint open sets containing $\pi(s)$ and $\pi(t)$ respectively? Thanks!

share|cite|improve this question
up vote 1 down vote accepted

I suspect that you’re getting a bit bogged down in the notation. It’s easier to see what’s going on if you temporarily forget about $X$ and work directly with $Y$. Your equivalence relation is chosen so that in effect you have a set $Y$ and a family $\mathscr{F}$ of functions from $Y$ to $\mathbb{R}$ such that whenever $x,y\in Y$ with $x\ne y$, there is some $f\in\mathscr{F}$ such that $f(x)\ne f(y)$. ($\mathscr{F}$ is simply the set your $\phi(f)$ for $f$ a continuous real-valued function on $X$.) Such a family is sometimes called a (point-)separating family of functions.

You’ve then taken $\mathscr{V}$ to be the coarsest topology making each $f\in\mathscr{F}$ continuous. To see that $\langle Y,\mathscr{V}\rangle$ is Hausdorff, let $x$ and $y$ be distinct points of $Y$; there is then some $f\in\mathscr{F}$ such that $f(x)\ne f(y)$. Let $U$ and $V$ be disjoint open intervals around $f(x)$ and $f(y)$; then $f^{-1}[U]$ and $f^{-1}[V]$ are disjoint open nbhds of $x$ and $y$ in $Y$.

share|cite|improve this answer
Thanks Brian, you make it look so simple. – Ashley Lin Oct 19 '11 at 19:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.