Timeline for Can an infinite number of mathematicians guess the number in a box with only one error?
Current License: CC BY-SA 3.0
13 events
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| Jun 20, 2022 at 10:55 | answer | added | Bennett McElwee | timeline score: 3 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 28, 2013 at 23:32 | vote | accept | user44653 | ||
| Dec 27, 2013 at 10:56 | comment | added | Asaf Karagila♦ | Ya know... this is the fourth question on this sort of puzzle over MO and MSE, and I still can't figure out what any of these have to do with the axiom of choice. I mean, sure, it's needed, but the axiom is also needed in constructing maximal ideals, ultrafilters, etc. | |
| Dec 27, 2013 at 4:10 | answer | added | Eric Naslund | timeline score: 16 | |
| Dec 26, 2013 at 23:30 | history | edited | user44653 | CC BY-SA 3.0 |
added 21 characters in body
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| Dec 26, 2013 at 23:18 | comment | added | user44653 | Good points about the countable sets of functions and thanks for the uncountability observation. In point of fact, the notion of the boxes being in "sequence" is unnecessary. A better way to describe the puzzle is probably to say there is a set of boxes cardinality $\alpha$, and to ask for the maximum cardinality $\beta$ of a set of strategies guaranteed to make at most $\gamma$ wrong guesses. But of course the countable/countable case is the most natural. | |
| Dec 26, 2013 at 9:12 | comment | added | Victor | This is an outcry of mathematical thought: "can an infinite number of mathematicians..." | |
| Dec 25, 2013 at 23:18 | comment | added | Dan Turetsky | I have a negative answer to the uncountable question. Notice that the behavior of a strategy doesn't depend on the value of the box it guesses at. If you're faced with uncountably many mathematicians, begin by placing 0 in all the boxes. By pigeon hole, there's some box that uncountably many of the mathematicians guess at. Adjust the value at that box to make most of them wrong. | |
| Dec 25, 2013 at 21:56 | comment | added | Noah Schweber | Just a quick observation: your argument about the case when $f$ is guaranteed to be computable isn't really about computability theory: a much stronger fact is true, namely if we are guaranteed that $f$ is in some pre-determined countable set of functions $\mathbb{N}\rightarrow\mathbb{R}$, then $k$ can be $\omega$. | |
| Dec 25, 2013 at 12:09 | history | edited | user44653 | CC BY-SA 3.0 |
edited title
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| Dec 25, 2013 at 8:49 | review | First posts | |||
| Dec 25, 2013 at 10:32 | |||||
| Dec 25, 2013 at 8:32 | history | asked | user44653 | CC BY-SA 3.0 |