Timeline for Given a sequence of reals, we can find a dense sequence avoiding it, but can we find one continuously?
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| Dec 15, 2016 at 18:48 | vote | accept | CommunityBot | moved from User.Id=44143 by developer User.Id=481663 | |
| Dec 15, 2016 at 11:31 | comment | added | Simon Henry | About the constructivity question at the end, in what sense your sequence is assumed to be injective ? $x_i = x_j \Rightarrow i=j$ ? or equivalently $i \neq j \Rightarrow x_i \neq x_j$ ? or the stronger form $i \neq j \Rightarrow x_i < > x_j $ ? (by $<>$ I mean $x < y$ or $y <x$) | |
| Dec 15, 2016 at 8:59 | comment | added | Pietro Majer | It seems interesting the problem with $\mathbb{R}^2$ instead of $\mathbb{R}$ (maybe easy?) | |
| Dec 15, 2016 at 8:41 | answer | added | Pietro Majer | timeline score: 6 | |
| Dec 15, 2016 at 7:54 | comment | added | user44143 | @PietroMajer, yes. | |
| Dec 15, 2016 at 7:31 | comment | added | Pietro Majer | By "topology of point-wise convergence", do you mean the topology induced on $S$ by the product topology of $\mathbb{R}^\mathbb{N}$? | |
| Dec 15, 2016 at 0:54 | answer | added | Joel David Hamkins | timeline score: 4 | |
| Dec 14, 2016 at 23:06 | comment | added | user44143 | @AndrejBauer, if you look at S', the set of all sequences, then it's clearly impossible. Define L_i = {s: f(s)(1) < s(i)} and R_i = {s: f(s)(1) > s(i)}, which are open sets with S' as the union of L_i and R_i. So either L_i = S' or R_i = S'. It is impossible to have L_i = S' and R_j = S', since sometimes s(i) < s(j). So either for all i, L_i = S', making f(s)(1) impossibly small; or for all i, R_i = S', making f(s)(1) impossibly large. (Credit here to an answer now deleted, and maybe there's a better way of writing this down that conveys the imagery of squeezing better.) | |
| Dec 14, 2016 at 22:11 | comment | added | Joel David Hamkins | You may also be interested in this related question: mathoverflow.net/a/47191/1946, concerning the impossibility of diagonalizing against countable sets of reals in a Borel way. | |
| Dec 14, 2016 at 21:55 | comment | added | Andrej Bauer | Why are you looking only at the injective sequences? | |
| Dec 14, 2016 at 21:54 | history | edited | user44143 | CC BY-SA 3.0 |
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| Dec 14, 2016 at 21:47 | comment | added | YCor | Simpler: do you know a way to map $s$ to an element $u(s)$ not in the range of $s$, in a "constructive" way (e.g., $u$ continuous, or even $u$ Borel)? it eventually seems to be equivalent to the problem, at least in its Borel/"constructive" version. | |
| Dec 14, 2016 at 21:45 | history | edited | YCor |
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| Dec 14, 2016 at 21:18 | history | asked | user44143 | CC BY-SA 3.0 |