Rats --- I did get there, but too late (reinvigorated by a post-prandial snooze, combined with further editorial sanction ...) To restore my bruised ego, I shall propose a supplementary. Denote by Lim(S) the set of accumulation points of set S ; and define the height h(S) to be the minimum n such that Lim^n(S) = { } is empty. Question: is h(S) necessarily finite for every grunch S ? Fred Lunnon On 10/28/16, Dan Asimov <asimov@msri.org> wrote:
Yes, that's it. Nice solving! Fred ... and Andy.
The puzzle was on an old Putnam exam, from those halcyon days when they didn't feel compelled to ask questions about the integer of the year A.D. in which the test was administered.
—Dan
On Oct 28, 2016, at 10:06 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thing I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0.
But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i.
On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org>> wrote: > Puzzle: Define a grunch to be any closed and bounded > subset of the real line R consisting entirely of rational numbers. > > True or false: There is a countable set of grunches such that > every grunch is a subset of one of them. > > Prove your answer.
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