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:


[https://topology.pi-base.org/spaces?q=Second%20Countable%20%2B%20Urysohn%20%2B%20~Metrizable](https://topology.pi-base.org/spaces?q=Second%20Countable%20%2B%20Urysohn%20%2B%20%7EMetrizable)


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.



Does second countable and functionally Hausdorff imply submetrizable?
**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.


