Timeline for Given a sequence of reals, we can find a dense sequence avoiding it, but can we find one continuously?

8 events

when toggle format	what		by	license	comment
Dec 16, 2016 at 0:58	comment	added	Joel David Hamkins		Yes, that's right. In fact it is very low in the Borel hierarchy.
Dec 16, 2016 at 0:57	comment	added	YCor		So for the reals, it follows that there is a Borel way of proceeding.
Dec 15, 2016 at 12:12	comment	added	Joel David Hamkins		@WłodzimierzHolsztyński The range of $f(s)$ is dense, since $y_k$ was chosen specifically to extend $u_k$, and so we've gotten into every basic open set. Fedor, my point is that if we use Cantor space or Baire space, then the answer is positive, even for the whole space of sequences; there is no need to restrict to the injective sequences. The main point is that Cantor diagonalization is a continuous process on these spaces.
Dec 15, 2016 at 6:47	comment	added	Włodzimierz Holsztyński		But $ran(f(s))$ are supposed to be dense. It seems that there is still some work to do.
Dec 15, 2016 at 6:21	comment	added	user44143		@FedorPetrov, he means that we can do it for the whole space of irrational sequences. It's a notably tweaked setup but an elegant solution to it.
Dec 15, 2016 at 6:17	comment	added	Fedor Petrov		But for reals and the whole space of sequences the answer is known (see my deleted comment, in which I forgot that we consider only injective sequences).
Dec 15, 2016 at 1:08	history	edited	Joel David Hamkins	CC BY-SA 3.0	added 16 characters in body
Dec 15, 2016 at 0:54	history	answered	Joel David Hamkins	CC BY-SA 3.0