Metrizability of Hausdorff continuous image of compact metric space

Question

Let $f$ be a continuous mapping of a compact metric space $(X, d)$ onto a Hausdorff space $(Y, \tau_1)$. Then $(Y, \tau_1)$ is compact and metrizable.

In one proof the following metric is constructed:

$d_1(y_1, y_2) = inf\{d(a, b) : a \in f^{-1}\{y_1\},\ b \in f^{-1}\{y_2\}\}$, for all $y_1$ and $y_2$ in $Y$.

I'm thinking about how to prove that triangle inequality holds for $d_1$, s.t. $d_1(x, z) \leq d_1(x, y) + d_1(y, z)$

And how about the case when triangle inequality does not hold?

Update: @daniel-schepler suggested to use metric $d_1(x, y) = inf\{d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)\}$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$.

Now we need to prove 3 property of metric: positivity, symmetry and triangle inequality. This is my sketch:

1. Positivity. Chose two points $x,y \in Y$ s.t. $x \neq y$. Singleton sets $\{x\}$ and $\{y\}$ are disjoint closed sets by Hausdorfness of $Y$ and their preimages are also disjoint closed in $X$ by continuity of $f$. $X$ is compact metric space by hypothesis and therefore a normal space. In normal space every two disjoint closed sets have disjoint open neighborhoods. Let $f^{-1}(x) \subseteq U$ and $f^{-1}(y) \subseteq V$ s.t. $U,V$ are open and $U \cap V = \varnothing$. Therefore there exists an $\varepsilon > 0$, s.t. $x^* \in B_\varepsilon(x^*) \subseteq U$ and $y^* \in B_\varepsilon(y^*) \subseteq U$ and distance between any two closed sets is strictly positive and $d_1$ is positive and zero iff $x = y$.

2. Symmetry. Clearly $d_1$ is symmetric.

3. Triangle inequality. For triangle inequality we can show that in the "worst" case $d_1(x, z) = d_1(x,y) + d_1(y, z)$.

Answer 1

No, this does not form a metric in general. For a specific counterexample, consider the case $X = [0, 1]^2$, $Y = \{ (x, y) \in [0, 1]^2 \mid x \le y \}$, and $f : X \to Y$ is defined by $f(x, y) = (xy, y)$. Then for all points in $Y$ with $y \ne 0$, the inverse image of $(x, y)$ is a single point $\{ (x/y, y) \}$; whereas the inverse image of $(0, 0)$ is $[0, 1] \times \{ 0 \}$. Thus, for example, in the induced $d_1$ on $Y$, we have $d_1((0, 0.1), (0.1, 0.1)) = d((0, 0.1), (1, 0.1)) = 1$. On the other hand, $d_1((0, 0.1), (0, 0)) = 0.1$ and also $d_1((0, 0), (0.1, 0.1)) = 0.1$.

You could try to fix this by considering paths, e.g. define $d_1(x, y)$ to be the infimum of values of the form $d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$. Then it would be straightforward to show this does satisfy the triangle inequality. However, then the tricky part would be to show that this infimum is strictly positive for $x \ne y$. I'm not sure off the top of my head how (or even if) that could be proved, though it seems clear that compactness would have to come into play in any proof of such a fact.

Answer 2

I'd go for a more general topology flavoured proof:

In a space $(X,\tau)$ we can define the notion of a network for the topology: A set of subsets $\mathcal{N}$ of $X$ is called a network for $(X,\tau)$ when for all $O \in \tau$, there is some $\mathcal{N'} \subseteq \mathcal{N}$ such that $O = \bigcup \mathcal{N}'$, or equivalently : $$\forall x \in O: \exists N_x \in \mathcal{N}: x \in N_x \subseteq O$$

Note that this is just like a base for the topology, except that in a base all elements are open, while here they can be any set. So in particular, a base is a special network, but e.g. the set of singletons is also a network, for any space $X$, so that if $X$ is any regular, countable but non-metrisable space, it has a countable network but not a countable base.

This notion is somewhat better behaved w.r.t. functions then bases are:

If $f: X \to Y$ is continuous and onto and $\mathcal{N}$ is a network for $X$, then $\{f[N]: N \in \mathcal{N}\}$ is a network for $Y$.

Proof: let $O$ be open in $Y$, let $y \in O$, then pick $x \in X$ with $f(x) = y$. $x \in f^{-1}[O]$, which is open, so there is some $N_x$ with $x \in N_x \subseteq f^{-1}[O]$. It follows that $y \in f[N] \subseteq O$. QED, as $y$ and $O$ were arbitary.

Note that this does not hold for bases, only when $f$ is an open map, because in a base the sets must be open.

Now a surprising fact, due to Arhangel'skij, who introduced this notion.

If $(X,\tau)$ is compact and Hausdorff, and has a network of infinite cardinality $\kappa$, then it also has a base of size $\kappa$.

If we assume this for now, we can show that the continuous Hausdorff image $Y$ of a compact metrisable space $X$ under a continuous map $f$ is also metrisable.

(the Hausdorffness of $Y$ is essential, or we take the co-finite or trivial topology on $X$, and $f$ the identity as a counterexample).

• $X$ is compact metrisable, so has a countable base.
• So $X$ has a countable network.
• So $Y$ has a countable network as $Y=f[X]$ and $f$ is continuous.
• Also, $Y$ is compact Hausdorff, so Arhangel'skij's theorem applies, and $Y$ has a countable base.
• As $Y$ is also $T_3$ (from compact Hausdorff again), the Urysohn metrisation theorem says that $Y$ is metrisable.

This does not construct a metric on $Y$, as your proof tries to, but it's nice and clean, and based on facts that are re-usable elsewhere.

Answer 3

For the sake of simplicity, here is a proof in the same vein which doesn't make use of networks.

By Urysohn's metrization theorem it suffices to show that $Y$ is second countable. Since the map $f$ is closed ($X$ being compact and $Y$ Hausdorff), it's natural to look for a countable base for the closed sets of $Y$.

(A base for the closed sets of $Y$ is a collection $\mathscr{C}$ of closed subsets of $Y$ such that any closed subset of $Y$ is the intersection of some collection of elements of $\mathscr{C}$; in other words, the collection $\{Y\setminus C: C\in\mathscr{C}\}$ is a base for the open sets of $Y$.)

To this end, let $D$ be a countable dense subset of $X$, and set $$ \mathscr{C}:=\left\{\bigcup_{k=1}^n f(\overline{B}(d_k,1/p)) : n,p\in\mathbb{N},d_1,\dots,d_n\in D\right\} $$ (where the notation $\overline{B}(x,r)$ denotes the closed ball of radius $r$ centered at $x\in X$); then $\mathscr{C}$ is countable, and each element of $\mathscr{C}$ is closed since $f$ is a closed map.

To show that $\mathscr{C}$ is a base for the closed sets of $Y$, choose any closed set $K\in Y$. Then $f^{-1}(y)$ and $f^{-1}(K)$ are disjoint compact sets of $X$, so $$ \delta:=\operatorname{dist}(f^{-1}(y),f^{-1}(K))>0 $$ Choose $p\in\mathbb{N}$ with $1/p < \delta/2$; since $f^{-1}(K)$ is compact there are finitely many points $d_1,\dots,d_n \in D$ such that $$ f^{-1}(K)\subseteq \bigcup_{k=1}^n \overline{B}(d_k,1/p) $$ and such that for each $k$ one has $\overline{B}(d_k,1/p)\cap K \neq \emptyset$. By the triangle inequality $$ \overline{B}(d_k,1/p)\cap f^{-1}(y) = \emptyset $$ for each $k$. In light of the above set relations, the surjectivity of $f$ implies that $$ K\subseteq \bigcup_{k=1}^n f(\overline{B}(d_k,1/p)) \subseteq Y\setminus \{y\} $$

Since $K$ and $y$ were arbitrary, $\mathscr{C}$ is a countable base for the closed sets of $Y$.