Er, spoiler and clarification below ... On 10/28/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
SPOILER below ...
On 10/27/16, Dan Asimov <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.
—Dan _______________________________________________
A bounded closed subset of reals is compact, so a grunch contains all its accumulation points, which must also be rational; conversely any bounded set of rationals containing its rational accumulation points is a grunch.
Suppose [H_i] is some denumerable sequence of grunches such that
for each grunch S there exists i such that S is a subset of H_i .
Let
G = {0} union {x_i} ,
where x_i is rational,
x_i notin Union_{j <= i} H_j ,
and
0 < x_i < x_{i-1} .
G contains its sole accumulation point 0 , so G is a grunch; however
for all i , G notin H_i
--- contradiction. Hence there is no such sequence [H_i] . QED
Fred Lunnon
Following a walk with the dog, not to mention a cryptic memo from Dan, it dawns on me that I failed to establish above that there exists some x_i with 0 < x_i < x_{i-1} and x notin Union_{j <= i} H_j . Firstly, it was unneccesary to take that union anyway: x notin H_i suffices. Now if every rational in (0, x_{i-1}) were a member of H_i , then all real numbers in that interval would be accumulation points, hence via compactness also members of H_i , which could therefore not be a grunch. By a second contradiction then, there must be some rational x_i satisfying the (weakened) constraint. Fred Lunnon