A topological space $\mathbf{X}$ is functionally Hausdorff, if for any two distinct $x, y \in \mathbf{X}$ there exists a continuous function $f_{xy} : \mathbf{X} \to [0,1]$ with $f(x) = 0$ and $f(y) = 1$.

A space $\mathbf{X} = (X,\tau)$ is submetrizable, if there exists a topology $\tau' \subseteq \tau$ such that $(X,\tau')$ is metrizable. Equivalently, if there is a continuous injection $\iota : \mathbf{X} \to \mathbf{X}'$ to some metric space $\mathbf{X}'$.

It is rather easy to see that any submetrizable space is functionally Hausdorff. I am wondering whether restricted to second-countable spaces, the converse might hold, too.

Failed solution attempts: My naive attempt to prove this was to pick a dense sequence $(a_n)_{n \in \mathbb{N}}$, and to consider the continuous map $F : \mathbf{X} \to [0,1]^\omega$ where $F(x)(\langle n,m\rangle) = f_{a_na_m}(x)$ for some tupling functions for unequal pairs. This map can fail to be injective, though: Take an uncountable space with a dense sequence of isolated points, pick the $f_{xy}$ suitably, and $$F[\mathbf{X}] = \{x \in [0,1]^\omega \mid \exists n \ \forall i \neq n \ x_n = 1 \wedge x_i = 0\} \cup \{0^\omega\}$$ is countable.

The initial topology induced by all $f_{xy}$ should be regular, and is nested between two countably-based topologies, but I do not see why it should be countably-based itself.

Searching on $\pi$-base for secound countable, functionally Hausdorff (aka Urysohn) but not metrizable spaces yields the following:


Of these examples most are just defined by adding open sets to a metrizable topology. The other two (irregular lattice topology and Roy's Lattice Subspace) are countable, hence the argument above with a total enumeration shows their submetrizability.

  • 1
    $\begingroup$ "Functionally Hausdorff" is in $\pi$-base, under the name "Urysohn". $\endgroup$ – Nate Eldredge Sep 5 '17 at 5:16
  • $\begingroup$ @NateEldredge Oh, I expected "Urysohn" to be a synonym for $T_{2.5}$. $\endgroup$ – Arno Sep 5 '17 at 6:42

Fact. Each second-countable functionally Hausdorff space is submetrizable.

This fact follows from a more general result:

Theorem. Each functionally Hausdorff space $X$ with hereditarily Lindelöf square $X\times X$ is submetrizable.

Proof. Denote by $\Delta$ the diagonal of the square $X^2:=X\times X$. For any distinct points $x,y\in X$ choose a continuous function $f_{x,y}:X\to[0,1]$ such that $f_{x,y}(x)=0$ and $f_{x,y}(y)=1$. Then $$U_{x,y}=\{(x',y')\in X\times X:f_{x,y}(x')<\tfrac12<f_{x,y}(y')\}$$ is an open neighborhood of $(x,y)$ in $X^2\setminus \Delta$. Since $X^2\setminus \Delta$ is Lindelof (by the hereditary Lindelofness of $X\times X$), the open cover $\{U_{x,y}:x,y\in X^2\setminus \Delta\}$ has a countable subcover $\{U_{x,y}\}_{(x,y)\in A}$ for some countable set $A\subset X^2\setminus\Delta$. Then the map $$f:X\to[0,1]^A,\;\;f:z\mapsto (f_{x,y}(z))_{(x,y)\in A}$$ is injective and hence $X$ is submetrizable (as $X$ admits a continuous injective map into the metrizable space $[0,1]^A$).

Corollary. Each functionally Hausdorff space with countable network is submetrizable.

  • $\begingroup$ The corollary is a very nice bonus! Are you aware of any published source to cite for this? Is it folklore or would be it be appropriate to attribute this to you? $\endgroup$ – Arno Sep 5 '17 at 10:23
  • $\begingroup$ @Arno I would consider this as a folklore. I admit that this argument has been used (many times) in topological literature, but know no exact reference at the moment. $\endgroup$ – Taras Banakh Sep 5 '17 at 10:37
  • $\begingroup$ I hadn't seen the concept of "network" before, interesting! It seems to be something like a neigbourhood basis, where we forget which points the basic neighbourhoods are attached to. $\endgroup$ – David Roberts Sep 5 '17 at 11:14
  • $\begingroup$ @DavidRoberts Network of the topology is "basis" without opennes (each open set is a union of sets from the network). I realizes that this definition resembles the definition of an ellipse as the "circle" inscribed into the square 3x4 :) $\endgroup$ – Taras Banakh Sep 5 '17 at 11:20

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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