# A totally disconnected non-metrizable space in which every open set is $F_{\sigma}$?

I am looking for a totally disconnected non-metrizable space in which every open set is open $$F_{\sigma}$$? For metric spaces, any countable or fnite set with discrete topology is totally D-disconnected, and only non metrizable example where every open set is open $$F_{\sigma}$$, I know is $$R^{\infty}$$ with weak topology which is connected. So any help on this is appreciated. Thanks.

$$\Bbb R$$ in the lower limit topology (which has as a base for its open sets all sets $$[a,b), a < b$$, also known as the Sorgenfrey line $$\Bbb S$$) is a classic example of such a space.

In a countable $$T_1$$-space, every set is an $$F_\sigma$$ set, right? So just take any non-metrizable totally disconnected countable $$T_1$$-space. For example choose a point $$p\in\beta\mathbb N\setminus\mathbb N$$ and let $$X=\mathbb N\cup\{p\}$$ topologized as a subspace of $$\beta\mathbb N$$.

For a simpler example, take $$X=(\mathbb N,\tau)$$ where $$\tau=\left\{U\subseteq\mathbb N:\text{ either }1\notin U\text{ or else }\sum_{n\in\mathbb N\setminus U}\frac1n\lt\infty\right\}.$$ Then $$X$$ is a countable zero-dimensional Hausdorff space which is not first countable. In particular, it is totally disconnected and not metrizable, and every subset is an $$F_\sigma$$.

Let $$T_{\Bbb Q}$$ be the usual topology on $$\Bbb Q$$ and identify $$\Bbb Z$$ to a point. So $$X=(\Bbb Q \setminus \Bbb Z)\cup \{p\}$$ with $$p\not \in \Bbb Q.$$

Details:

(1). Let $$T_X$$ be the topology on $$X.$$ If $$p\not\in U\subset X$$ then $$U\in T_X$$ iff $$U\in T_{\Bbb Q}.$$ If $$p\in U\subset X$$ then $$U\in T_X$$ iff $$(U\setminus \{p\})\cup \Bbb Z\in T_{\Bbb Q}.$$

$$X$$ is a countable $$T_1$$ space so every subset is $$F_{\sigma}.$$ (BTW, $$X$$ is completely regular.)

(2). For $$z\in \Bbb Z$$ and $$q\in \Bbb Q\cap (z,z+1)$$ let $$r_q=\min (q-z,z+1-q).$$ Then $$B(q)=\{\Bbb Q\cap (-t+q,t+q):t\in (0,r_q)\setminus \Bbb Q\}$$ is a local base at $$q,$$ and each member of $$B(q)$$ is open-and-closed.

(3). Let $$J=\{(t(z))_{z\in \Bbb Z}: \forall z\in \Bbb Z\,(\,t(z)\in (0,1/2)\setminus \Bbb Q\,)\}.$$

For $$\tau=(t(z))_{z\in \Bbb Z} \in J$$ let $$f(\tau)=\{p\}\cup [(\Bbb Q\setminus \Bbb Z)\cap (\;\cup_{z\in \Bbb Z}(-t(z)+z,t(z)+z)\;)].$$ Then $$B(p)=\{f(\tau): \tau \in J\}$$ is a local base at $$p$$ and every member of $$B(p)$$ is open-and-closed.

(4). $$X$$ is not metrizable because it is not 1st-countable:

Let $$V=\{U_x:x\in \Bbb Z\}$$ be any countable family of nbhds of $$p.$$ For each $$x\in \Bbb Z$$ take $$\tau_x=(t_x(z))_{z\in \Bbb Z}\in J$$ such that $$f(\tau_x)\subset U_x,$$ and let $$g(x)=\frac {1}{2}t_x(x).$$ Then $$G=(g(x))_{x\in \Bbb Z}\in J$$ so $$f(G)$$ is a nbhd of $$p$$. And no member of $$V$$ is a subset of $$f(G).$$