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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 $fāˆ˜f^\omega āŠ† f^\omega$ (which is implied by my condition above. This becomes clearer if you write it as $f^\omega = \bigcup_{nāˆˆā„•} f^{n\downarrow}$ 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