Is R with $j_d$ topology totally disconnected? Let consider the topological space $R_j=(\mathbb{R},j_r)$ where $j_r$ is generated by right side open intervals, i.e $[a,b)$ for $a,b \in \mathbb{R}$; note that this topology includes the euclidean topology.
$R_j$ is not a connected space, because given $x \in \mathbb{R}$ we have that $(-\infty,x)$ and $[x,+\infty)$ provide a disconnection or $R_j$.
Given the arbitrariness in choosing $x$, is it possible that $R_j$ is totally disconnected? Let consider a certain $y:x<y<+\infty$; I can progressively reduce the open interval containing $x$ $$[x...)...)...)...)...y)$$ so that the connected component which contains $x$ is (strictly) included in $[x,x+\varepsilon)\;\forall\varepsilon>0.$ if we suppose there is a certain limit in this reduction, then we would have to accept that $[x-\varepsilon /2,x+\varepsilon /2)$ is a connected component, which is a contradiction with the beginning of this discussion. 
What can we conclude, from this?
 A: This space is usually called the Sorgenfrey line. You can also prove directly that it’s totally disconnected. If $A\subseteq\Bbb R$ has at least two points, choose $x,y\in A$ with $x<y$; then the sets$A\cap(\leftarrow,y)$ and $A\cap[y,\to)$ separate $A$, so $A$ is not connected.
In fact a stronger statement is true, one that’s actually even easier to prove: the Sorgenfrey line is $0$-dimensional, meaning it has a base of clopen sets, namely, the base $\big\{[x,y):x,y\in\Bbb R\text{ and }x<y\big\}$. The proof that every $0$-dimensional space is totally disconnected is just a generalization of the proof that I gave above showing that the Sorgenfrey line is disconnected.
A: Your suspicion that $R_j$ is totally disconnected is correct. Note that since $R_j$ is strictly finer than the standard real topology, then it has fewer connected sets (or at least no extras). Thus, our only connected candidates are intervals, rays, singletons, the whole line, and the empty set. Well, you've already shown that the whole line isn't connected, and a similar approach will show that no interval or ray is connected. Thus, the only non-empty connected sets are singletons, as desired.
