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". $\endgroup$ Sep 5, 2017 at 5:16
  • $\begingroup$ @NateEldredge Oh, I expected "Urysohn" to be a synonym for $T_{2.5}$. $\endgroup$
    – Arno
    Sep 5, 2017 at 6:42

1 Answer 1


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, 2017 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. $\endgroup$ Sep 5, 2017 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, 2017 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 :) $\endgroup$ Sep 5, 2017 at 11:20

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