Timeline for Does Arzelà-Ascoli require choice?
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| Sep 12, 2015 at 18:32 | comment | added | Cloudscape | In fact, I think I see a proof for the sequential compactness of $[0, 1]^\omega$ without choice, by using the same method as in the paper. We first choose a subsequence convergent in the first dimension, then the next and so on, always choosing the smallest accumulation point w.r.t some suitable total order. Then we diagonalise and obtain a sequence convergent in $[0, 1]^\omega$ (it always converges into the respective finite "box"). | |
| Sep 12, 2015 at 16:10 | comment | added | Cloudscape | Should we move this conversation to chat? BTW: I am just now preparing an update of the paper with several inaccuracies fixed (EDIT: Only minor stuff). I will upload this later in the evening. | |
| Sep 12, 2015 at 15:46 | comment | added | Nate Eldredge | Well, compactness of $[0,1]^\omega$ doesn't need any choice. See here. I am still not sure about sequential compactness. | |
| Sep 12, 2015 at 15:30 | comment | added | Nate Eldredge | Oh, it's not Lipschitz as it stands, because if $f(1/n)=1/n$ and $f(1/(n+1))=0$ then the output changes by $1/n$ for an input change of $\approx 1/n^2$. But maybe mapping $1/n$ to $u(n)/n^2$ or something like that would fix it. As for getting back to convergence in $[0,1]^\omega$, I think that is clear since a uniformly convergent sequence converges pointwise. I should say that I'm also not completely sure how much choice is needed for "$[0,1]^\omega$ is sequentially compact". | |
| Sep 12, 2015 at 15:14 | comment | added | Cloudscape | Seems like a sensible idea! If my stuff is wrong, I want to know! I wouldn't like to spread wrong stuff, so I don't take it personally or something. The piecewise linear function should be Lipschitz continuous with Lipschitz constant $1$; this would give equicontinuity, and boundedness is clear. It remains to prove that convergence in the continuous functions implies convergence in $[0, 1]^\omega$. | |
| Sep 12, 2015 at 15:01 | comment | added | Nate Eldredge | I haven't thought out the details, but my idea was to identify $u \in [0,1]^\omega$ with a continuous function that maps $1/n$ to $u(n)/n$, perhaps in a piecewise linear fashion. It is entirely possible that I am missing something, so I should probably think it through before saying any more :-) | |
| Sep 12, 2015 at 14:59 | comment | added | Cloudscape | I think you still need choice, because the additional point in "my theorem" is that the points "weigh" differently; in $[0,1]^\omega$ all the coordinates are equal w.r.t. the topology, but in Arzela-Ascoli they become weaker and weaker as the induction proceeds. | |
| Sep 12, 2015 at 14:55 | comment | added | Cloudscape | Could you sketch a proof of "my theorem $\Rightarrow$ $[0, 1]^\omega$ is seq. compact"? | |
| Sep 12, 2015 at 14:49 | comment | added | Nate Eldredge | Hmm, this is surprising. It seems to me that using your Theorem, I could prove that $[0,1]^\omega$ is sequentially compact, and I was under the impression that this wasn't provable without some form of choice. I have to do some reading, I guess. | |
| Sep 12, 2015 at 13:56 | review | Late answers | |||
| Sep 12, 2015 at 14:02 | |||||
| Sep 12, 2015 at 13:41 | review | First posts | |||
| Sep 12, 2015 at 14:55 | |||||
| Sep 12, 2015 at 13:38 | history | answered | Cloudscape | CC BY-SA 3.0 |