# Semi-metrizable spaces with countable chain condition

Note that $X$ is semi-metrisable iff $X$ is first countable and semi-stratifiable.

## Definition

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.

• What is the definition of a semi-metrizable space? – Taras Banakh Apr 20 '17 at 21:35
• @TarasBanakh see the revised Question. – Paul Apr 21 '17 at 4:23
• "Countable chain condition" is short for CCC???? – bof Apr 21 '17 at 10:12
• Is it: $\ \left(\forall_n (x\in g(n\ x_n)\lor x_n\in g(n\ x))\right)\ \Rightarrow\ (\lim x_n = x)\$ ? – Włodzimierz Holsztyński Apr 21 '17 at 17:36
• @WłodzimierzHolsztyński Yes – Paul Apr 22 '17 at 0:14

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 $\min\{x_V,x_U\}$.