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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.

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$\Bbb R$ 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.

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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 $F_\sigma$.

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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 $f(G).$

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