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 Dec 29 accepted Scott continuity on powerset Dec 28 comment Scott continuity on powerset @MphLee: yes, sorry. It is idempotent iff $fāf^\omega ā f^\omega$ (which is implied by my condition above. This becomes clearer if you write it as $f^\omega = \bigcup_{nāā} f^{n\downarrow}$ with $f^{n\downarrow} = \bigcup_{i\leq n} f^i$.) Dec 28 comment Scott continuity on powerset Got it, thank you! And $x_d$ can be an upper bound of the finite set $x$ by definition of directed set (not sure why you talked about compact elements). I'll accept the answer in a while if nobody answers 2. Dec 28 asked Scott continuity on powerset Dec 16 awarded Caucus Apr 8 awarded Critic Mar 17 awarded Yearling Feb 21 answered Sum of two periodic functions is periodic? Jul 10 awarded Citizen Patrol Mar 17 awarded Yearling Jul 31 accepted Bounded simulations between Bernoulli distributions Jul 31 comment Bounded simulations between Bernoulli distributions @did I definitely got something, I'll accept it. Jun 8 awarded Constituent Jun 8 awarded Caucus Jun 6 awarded Scholar Jun 6 accepted Enumerating all $x$ such that $b^n$ divides $x^2-x$ Jun 5 revised Enumerating all $x$ such that $b^n$ divides $x^2-x$ Small mistakes. Jun 5 comment Enumerating all $x$ such that $b^n$ divides $x^2-x$ Thanks, notably for the fact that there is at most one solution for each $S_x$. Jun 5 suggested approved edit on Enumerating all $x$ such that $b^n$ divides $x^2-x$ Jun 5 revised Enumerating all $x$ such that $b^n$ divides $x^2-x$ b=14