Timeline for Is there a continuous surjection $\omega^\omega\to \mathbb{R}$?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2019 at 1:26 | comment | added | Robin Saunders | Put another way: the corresponding function from $\omega^\omega \rightarrow \mathbb{R}$ is discontinuous at every eventually-zero sequence, and these form a dense set. | |
| Oct 28, 2019 at 1:21 | comment | added | Robin Saunders | Note, though, that if $a_n$ were all $\ge 1$ then you could choose any $k$ you liked and $[a_0; a_1, ..., a_n + k]$ would always converge to the true value (I'm used to choosing $k = 0$). But if $a_n$ are all $= 0$ then the "convergents" oscillate between $k$ and $1/k$, and only converge if you happened to choose $k = 1$. This is surely evidence that $1$ is "the right answer", especially when combined with the observation about solutions to $x = 0 + 1/x$, but it doesn't feel watertight. | |
| Oct 28, 2019 at 1:21 | comment | added | Robin Saunders | Sure, you could take that as a definition which agrees with the usual one (in terms of strictly positive sequences) where the latter exists. | |
| Oct 28, 2019 at 0:25 | comment | added | Noah Schweber | @RobinSaunders Isn't it enough to just look at the sequence $$a_1+1, a_1+{1\over a_2+1}, a_1+{1\over a_2+{1\over a_3+1}}, ...$$ to handle things? Or am I missing something? | |
| Oct 28, 2019 at 0:04 | comment | added | Robin Saunders | Final(?) twist: x = 1/x actually has two solutions! At least -1 is also an integer, so in some sense you can get each rational "one way or the other". | |
| Oct 27, 2019 at 23:58 | comment | added | Robin Saunders | Good to know I wasn't missing something incredibly obvious! | |
| Oct 27, 2019 at 23:58 | comment | added | Noah Schweber | @RobinSaunders Haha, overlapping comments. I think the state of affairs is: it can make sense, and did to me back when I wrote this post, but current-me was hopelessly confused. | |
| Oct 27, 2019 at 23:56 | comment | added | Robin Saunders | Ok, I think I see now. Of course, n + 1/(0 + 1/x) = n + x, which takes care of sequences which are not eventually-zero, but might not be eventually-nonzero either. Then for a sequence of all zeroes, solving x = 0 + 1/x gives x = 1. I found this unfamiliar because I'm used to thinking of a partial fraction as just the limit of its convergents. | |
| Oct 27, 2019 at 23:29 | comment | added | Robin Saunders | I'm familiar with continued fractions which terminate, but not with ones containing zeroes. I can see how to interpret the latter when the sequences are eventually-nonzero, but find it non-obvious that ones built from eventually-zero sequences even make sense. (Edit: I'd be grateful for a reference on this, and didn't mean to ask for a direct explanation. Searching for "continued fractions" "containing zeroes" and the like hasn't turned up anything useful so far.) | |
| Oct 27, 2019 at 23:10 | comment | added | Robin Saunders | Slight quibble: continued fractions only get you irrational numbers (the map in question is a homeomorphism). There are various simple tricks to get you the rest, see mathoverflow.net/questions/112127/… | |
| Sep 12, 2017 at 14:29 | comment | added | Nate Eldredge | The other one that's often useful is that there is a continuous embedding from any separable metrizable space $X$ to the Hilbert cube $[0,1]^\omega$. (Pick a countable dense subset $\{x_n\} \subset X$ and a metric $d$, and consider the map $x \mapsto (d(x_1, x), d(x_2,x), \dots)$.) | |
| Sep 12, 2017 at 14:21 | comment | added | Asaf Karagila♦ | Well. Also $\Bbb R$ is connected, so any continuous function to the Baire space is constant. | |
| Sep 12, 2017 at 14:00 | vote | accept | Dominic van der Zypen | ||
| Sep 12, 2017 at 13:59 | comment | added | Dominic van der Zypen | Brilliant, thanks @NoahSchweber also for the further comments!! | |
| Sep 12, 2017 at 13:38 | history | answered | Noah Schweber | CC BY-SA 3.0 |