Note that $X$ is semi-metrisable iff $X$ is first countable and [semi-stratifiable.](https://mathoverflow.net/questions/267068/is-every-semi-stratifiable-space-omega-monolithic)


Definition
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A topological space $(X,\tau)$ is called **semi-metric** if there exists a function $g:\omega\times X\to\tau$ such that:


1. for any point $x$ of $X$ holds $\{x\}=\bigcap\_{n\in\omega} g(n,x)$;
2. for any point $x$ of $X$ and a sequence $\{x\_n\}$ of $X$, if $x \in g(n,x\_n)$ or $x\_n \in g(n, x)$ for each $n$, then $x\_n \to x$.


Is every semi-metrizable space with countable chain condition (short for CCC)
separable? Or is there a counterexample?


Thanks very much.



Semi-metrizable spaces with countable chain condition
There exists a counterexample, that is, a non-separable semi-metrizable space with ccc, which can be constructed as follows. Consider the countable power $\mathbb Z^\omega$ of the group of integers. For every $n\in\omega$ consider the "upper cone" $$V\_n=\{x\in\mathbb Z^\omega:\forall i<n\;x(i)=0\mbox{ and }\forall i\ge n\;x(i)\ge0\}$$ and its symmetrization $X\_n:=V\_n\cup(-V\_n)$.


On the group $\mathbb Z^\omega$ consider the topology $\tau$ consisting of sets $U\subset\mathbb Z^\omega$ such that for every $x\in U$ there exists $n\in\omega$ such that $x+X\_n\subset U$. 


The space $X=(\mathbb Z^\omega,\tau)$ (which is an Abelian quasi-topological group) has the required properties: it is semi-metrizable, non-separable and has ccc. The semi-metrizability of $X$ is witnessed by the function $g(n,x)=x+X\_n$. The non-separability of $X$ follows from the fact that for every countable subset $A\subset \mathbb Z^\omega$ there is a function $f\in\mathbb Z^\omega$ such that $f\notin\bigcup\_{a\in A}(a+X\_0)$ and hence $A\cap(f+X\_0)=\emptyset$. 


To see that $X$ has ccc, take an uncountable family $\mathcal U$ of non-empty open sets. We can assume that each $U\in\mathcal U$ is of basic form $x\_U+X\_{n\_U}$ for some $x\_U\in\mathbb Z^\omega$ and $n\_U\in\omega$. By the Pigeonhole Principle, there exists $n\in\omega$ such that the subfamily $\mathcal U\_n=\{U\in\mathcal U:n\_U=n\}$ is uncountable. Since the set $\mathbb Z^n$ is countable, we can apply the Pigeonhole Principle once more and find an element $x\in\mathbb Z^n$ such that the set $\mathcal U\_{n,x}=\{U\in\mathcal U\_n:x\_U|n=x\}$ is uncountable. Finally, take any two distinct sets $U,V\in\mathcal U\_{n,x}$ and observe that the intersection $U\cap V$ is not empty as it contains the functions $\max\{x\_V,x\_U\}$ and $\min\{x\_V,x\_U\}$.


