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Timeline for Shape of long sequences in C(ω_1)

Current License: CC BY-SA 2.5

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Apr 12, 2010 at 18:54 vote accept David R. MacIver
Apr 12, 2010 at 12:42 history edited Joel David Hamkins CC BY-SA 2.5
Fixed definition of r, using separability.
Apr 12, 2010 at 12:34 comment added Joel David Hamkins And I shall fix the typos presently.
Apr 12, 2010 at 12:33 comment added David R. MacIver Oh, yes, of course. I'm sorry. Misreading.
Apr 12, 2010 at 12:30 comment added Joel David Hamkins David, my proof wasn't claiming that the r_beta are simultaneously constant from gamma out to omega_1, but only that r_beta is constant on the interval [gamma,beta]. In the example of your comment, gamma=0 works for this, since every r_alpha is constant from [0,alpha] in your example. And the sequence I build would have s=1. So, the sequence provided by my solution has value 1 from gamma=0 onwards, which works for this example.
Apr 12, 2010 at 8:21 comment added David R. MacIver Actually... I'm starting to have serious doubts about the core idea of this proof. It doesn't seem to use anything about the $r_\alpha$ other than that they're continuous, and it's certainly not the case that every $\omega_1$ sequence of continuous functions is mutually constant after some point. Why, for example, would this not work with the following: Let $x_\alpha = 1_{[0, \alpha]}$ and let $r_\alpha = x_\alpha$. Obviously this isn't a bad sequence, but there's no common point above which the $r_\alpha$ are all constant.
Apr 12, 2010 at 8:14 comment added David R. MacIver Another wrong but fixable detail: I don't believe it works to use $r_\gamma$ up to and including stage $\gamma$. The problem is that later elements of the sequence may add variation before $\gamma$ even while being well behaved after it. However, $C([0, \gamma])$ is separable, so has no bad sequences, so we can still find a center that works for all of $[0, \gamma]$ and then use the constant center after that as per your argument.
Apr 12, 2010 at 7:49 comment added David R. MacIver Oh, by the way, it doesn't actually damage the proof, but your interval is slightly wrong. I think you actually mean $(\gamma, \beta]$. When $\beta$ is a successor ordinal you can't guarantee that there are any smaller $\gamma$ with value close to it.
Apr 12, 2010 at 7:44 comment added David R. MacIver Thanks. I'm pretty sure I should have spotted something like that (it looks very similar to the proof which got me to this problem in the first place), but I'm not sure I would have! What's the etiquette/convention for citing a mathoverflow answer in a paper?
Apr 12, 2010 at 7:42 vote accept David R. MacIver
Apr 12, 2010 at 8:19
Apr 12, 2010 at 7:42 vote accept David R. MacIver
Apr 12, 2010 at 7:42
Apr 12, 2010 at 1:20 history answered Joel David Hamkins CC BY-SA 2.5