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?