Is $\mathbb R_l$ totally disconnected? We know $\mathbb R_l $ is finer than $\mathbb R$ and intervals and one point sets are only connected subsets of $\mathbb R$ .hence only possible connected sets in $\mathbb R_l$ is intervals and one point set .as intervals are seperated by $(-\infty ,a),[a,\infty)$ in $\mathbb R_l$. Therefore one point sets are only connected sets in $\mathbb R_l$.

Am I wrong?

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    $\begingroup$ It seems fine to me. $\endgroup$ – pisco Nov 12 '17 at 5:29
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    $\begingroup$ Correct. The Sorgenfrey line has a base (basis) of open-and-closed sets. $\endgroup$ – DanielWainfleet Nov 12 '17 at 10:57

Direct proof. If distinct a,b in connected K then, wlog
a < b and (-oo,b), [b,oo) disconnects K by two open sets.

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