Apparently it's a standard result that, although not every (Hausdorff) quotient of a metric space is metrizable, it always is metrizable when the space being quotiented is compact. Alas, I can't find a proof - neither in textbooks nor one of my own. Can anyone give a hint, or a reference (or a proof, of course)?

Question 2: Is every quotient of a compact metric space (i.e. whether this quotient is Hausdorff or not) second-countable?


2 Answers 2


Let $X$ be a compact metric space and $f:X\to Y$ a quotient map, where $Y$ is Hausdorff. For each $y\in Y$ let $F_y=\{x\in X:f(x)=y\}$, the fibre of $f$ over $y$; $\{F_y:y\in Y\}$ is a partition of $X$ into closed sets. Let $\mathscr{B}$ be a countable base for $X$; without loss of generality we may assume that $\mathscr{B}$ is closed under finite unions.

Let $K\subseteq X$ be closed; then $K$ is compact, so $f[K]$ is compact and hence closed (since $Y$ is Hausdorff). Thus, $f$ is a closed map, and for each closed $K\subseteq X$ the set

$$f^{-1}\big[f[K]\big]=\bigcup_{y\in f[K]}F_y$$

is closed in $X$. It follows that if $U\subseteq X$ is open, then so is

$$\widehat U=\bigcup\{F_y:y\in Y\text{ and }F_y\subseteq U\}\;.$$

In words, the outer saturations of closed subsets of $X$ are closed, and the inner saturations of open subsets of $X$ are open.


$$\mathscr{B}_Y=\left\{f[\widehat B]:B\in\mathscr{B}\right\}\;;$$

$\mathscr{B}_Y$ is a countable family of open subsets of $Y$. Let $y\in Y$, and let $U$ be an open nbhd of $y$ in $Y$. Then $f^{-1}[U]$ is an open nbhd of $F_y$ in $X$. For each $x\in F_y$ there is a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq f^{-1}[U]$. $\{B_x:x\in F_y\}$ is an open cover of $F_y$, $F_y$ is compact, and $\mathscr{B}$ is closed under finite unions, so there is a $B\in\mathscr{B}$ such that $F_y\subseteq B\subseteq f^{-1}[U]$. Then $F_y\subseteq\widehat B\subseteq f^{-1}[U]$, so $y\in f[\widehat B]\subseteq U$, where $f[\widehat B]\in\mathscr{B}_Y$. Thus, $\mathscr{B}_Y$ is a countable base for the compact Hausdorff space $Y$, which is therefore metrizable.

The answer to the second question is no. Let $X=[0,1]$, $Q=[0,1]\cap\Bbb Q$, and let $Y=X/Q$. If $f$ is the quotient map, it’s not hard to check that $U\subseteq Y$ is open iff $f^{-1}[U]$ is an open nbhd of $Q$ in $[0,1]$. Thus, $Y\setminus\{f(x)\}$ is open in $Y$ for each $x\in X\setminus Q$. If $Y$ were second countable, the singleton $f[Q]$ in $Y$ would therefore be a $G_\delta$ in $Y$, and $Q$ would be a $G_\delta$ in $X$, contradicting the Baire category theorem.


Here is a somewhat high-tech proof for the first question that I'm fond of. First, note that if $K$ is a compact Hausdorff space, then $K$ is metrizable iff the space $C(K)$ of continuous real-valued functions on $K$ is separable (with respect to the sup norm). Indeed, if $K$ is metrizable iff $K$ embeds in the space $[0,1]^\mathbb{N}$, and $K$ embeds in $[0,1]^\mathbb{N}$ iff there are countably many elements of $C(K)$ that separate points of $K$. If $C(K)$ is separable, you can find countably many such elements by taking a countable dense subset. Conversely, if countably many functions separate points, then by the Stone-Weierstrass theorem the set of all polynomials in these functions with rational coefficients is dense in $C(K)$, so $C(K)$ is separable.

Now we simply observe that if $X$ is a compact metric space and $K$ is a Hausdorff quotient of $X$, the quotient map $X\to K$ induces an isometric embedding $C(K)\to C(X)$. Since $X$ is metrizable, $C(X)$ is separable. Since any subspace of a separable metric space is separable, it follows that $C(K)$ is separable, and hence $K$ is metrizable.

  • $\begingroup$ In general, if $f:X \to Y$ is continuous and onto, $\textrm{nw}(Y) \le \textrm{nw}(X)$, where nw denotes the network weight. For compact Hausdorff spaces $\textrm{nw}(X)=w(X)$ so if a quotient of a compact metrisable space is Hausdorff it is metrisable by Urysohn, essentially. $\endgroup$ Sep 9, 2019 at 21:05

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