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Jan 15, 2012 at 4:38 history undeleted S. Carnahan
Dec 23, 2011 at 17:44 history deleted Andrés E. Caicedo
Kevin Buzzard
Andy Putman
Feb 14, 2010 at 1:57 comment added Pete L. Clark @JDH: We are all finding our way in this new social environment, so what I say is tentative, but: it seems to me that if the questioner wants a question to be closed, it should be -- and stay -- closed. I think Harry is right when he says that anyone who is interested in the question can easily ask a new (and, one hopes, improved) version of the question. Please go ahead and ask a new question if you like: the topic seems interesting to me too.
Feb 14, 2010 at 1:30 comment added Harry Gindi I've got a lot of work and an exam this week, so if you can think of an interesting way to ask it, feel free to re-ask it yourself in another question. I say this mainly because I don't know how to fix the question (although admittedly I haven't given it much thought).
Feb 14, 2010 at 1:26 comment added Harry Gindi @Joel: When I figure out how to edit the question to make it better, I'm just going to ask it as a new question.
Feb 14, 2010 at 0:58 comment added Joel David Hamkins I voted to re-open, as a way to encourage you to edit the question. I continue to think that there is something interesting here.
Feb 13, 2010 at 12:19 history closed Harry Gindi
Joel David Hamkins
Bjorn Poonen
Pete L. Clark
Tom Leinster
no longer relevant
Feb 13, 2010 at 6:38 comment added Harry Gindi Well, we can always reopen it later. I'd rather have some time to think about it.
Feb 13, 2010 at 6:28 comment added Bjorn Poonen @fpqc: I too did what you asked, but I agree with Joel that there is something interesting here. Maybe it could be reworded somehow to encourage the kind of answers you are hoping for.
Feb 13, 2010 at 6:21 comment added Harry Gindi I guess I didn't put enough thought into the question, as Pete revealed to me via the Socratic method.
Feb 13, 2010 at 6:20 comment added Joel David Hamkins Well, I did what you asked, but actually, I find it interesting.
Feb 13, 2010 at 6:16 comment added Harry Gindi I voted to close as well.
Feb 13, 2010 at 6:15 history made wiki Post Made Community Wiki by Harry Gindi
Feb 13, 2010 at 6:15 comment added Harry Gindi Could you guys close this? I know at least Joel and Pete can vote to close.
Feb 13, 2010 at 6:15 comment added Bjorn Poonen OK, here are 2^{2^{\aleph_0}} examples: Let s be any field automorphism of C, and consider {s \circ f \circ s^{-1} : f is continuous in the usual sense}!
Feb 13, 2010 at 6:11 comment added François G. Dorais Example machine: Take X to be the set of all continuous function for a topology on C with respect to which addition and multiplication are continuous...
Feb 13, 2010 at 6:09 comment added Pete L. Clark @Harry: I am suggesting an answer to the question posed in the first paragraph, or rather suggesting a strategy for you to answer the question yourself. Contrary to your latest remark, it doesn't sound to me like you have worked through a counterexample. I encourage you to do this.
Feb 13, 2010 at 5:58 answer added Joel David Hamkins timeline score: 2
Feb 13, 2010 at 5:57 history edited Harry Gindi CC BY-SA 2.5
added 256 characters in body
Feb 13, 2010 at 5:56 comment added Harry Gindi @Bjorn, I'm not really looking for counterexamples, I'm looking for examples.
Feb 13, 2010 at 5:55 comment added Harry Gindi @Pete: It's obviously false in general. I was really wondering about any interesting cases where it is true. I just stated it the way I did because it's more interesting when stated that way.
Feb 13, 2010 at 5:55 comment added Bjorn Poonen If X does not contain the identity function, then X is a counterexample. For example, the set of constant functions is a counterexample. (Or do you interpret "closed under composition" as including 0-fold compositions, which would give you the identity for free?)
Feb 13, 2010 at 5:52 comment added Harry Gindi @FGD: If the topology is not the same on both sides, then the problem is pretty trivial.
Feb 13, 2010 at 5:49 comment added Pete L. Clark What if you take X to be the set of polynomial functions? Then the question asks whether or not there is a larger class of functions (with the closure properties you mention) which is continuous with respect to the Zariski (= cofinite, here) topology. There are lots of non-polynomial Zariski-continuous functions: e.g., note that pulling back the cofinite topology by any permutation of C gives the cofinite topology. Have you tried working this out?
Feb 13, 2010 at 5:45 comment added François G. Dorais Do you want the same topology at both ends, or just on the domain side?
Feb 13, 2010 at 5:38 history asked Harry Gindi CC BY-SA 2.5