TITLE: Is R with $j_d$ topology totally disconnected?
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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.