When is double negation Scott continuous?
Let $L$ be the frame of opens of a $T_0$-space $X$. We prove that if $X$ is sober and $T_1$, then the double negation nucleus on $L$ is Scott continuous iff $X$ is discrete. It follows that if, in addition, $X$ is compact then double negation is Scott continuous iff $X$ is finite. We show that both the sober and $T_1$ assumptions are essential, and generalize the above results to all boolean nuclei on $L$. A pointfree characterization of when $X$ is sober and $T_1$ is also given by proving that it is equivalent to Scott continuous nuclei on $L$ being closed.
math.LO↗