Problem: Give $\mathbb{R}$ the floor topology. Is $\mathbb{R}$ totally disconnected? Discrete?
As per the definition of floor topology, if we let $a,b \in \mathbb{R}$, then the floor topology basis will be $$\beta_l = \{ [a,b), \text{ where } a< b \in \mathbb{R}\}$$
Here, $[a,b)$ is both open and closed in $\beta_l$.
Also, from the definition of totally disconnected space (i.e. a space $X$ is said to be totally disconnected iff the only connected subsets of $X$ are singletons), and definition of subspace topology, we can say that the topology on $\mathbb{R}$ will be (say) $T_R = \{V \subset \mathbb{R} \mid \exists U \in \beta_l \text{ s.t. } V = U \cap \mathbb{R}\}$. Now, if we take a collection $U$, i,e, an element of $\beta_l$ (floor or lower limit topology), then $V = U \cap \mathbb{R}$ will give us all the singletons, and so the only connected subsets will be singletons. Therefore, $\mathbb{R}$ is totally disconnected.
Now, let $\{a\}$ be open for any $a \in \mathbb{R}$, therefore $[x,a)$ is open in $\beta_l$ for some $x \in \mathbb{R}$. Then, $[x,a) \cup \{a\}$ is open in $\beta_l$, but $[x,a) \cup \{a\} = [x,a]$ is not open in $\beta_l$. Therefore, $\beta_l$ is not discrete. Therefore it is not discrete but it is totally disconnected. Proved.
I hope this looks good. I am unsure about whether I used subspace topology properly to show that its totally disconnected. I have an idea but I don't think I showed n explained it well (something which my prof expects me to do all the time :P). Hence, I need someone to verify this. Also, let me know if it needs a better notation and style. Appreciate your help!!
Best Answer
Your effort is a mess, I'm sorry to say.
Note that any basic open set $[a,b)$ where $a< b$ is open and closed at the same time (clopen).
This means that if $A \subseteq \Bbb R$ is any subset with two points or more in it (say $a_1 < a_2$ are points of $A$) then $[a_1,a_2)$ and $A\setminus [a_1,a_2)$ are both open in $A$, by definition they are a partition of $A$ and both are non-empty (the two points are in different sets). So $A$ is disconnected. This argument works in general for any Hausdorff space with a clopen base. So $\Bbb R$ in the "floor topology" (i.e. the Sorgenfrey line) is totally disconnected.
It is not discrete as all basic open sets are infinite and so all non-empty open sets are infinite and so it has no isolated points a fortiori.