[math-fun] continuum version of vector puzzle
Now consider the set of all vectors in n dimensions, each of whose entries lies in the real closed interval [-1,+1]. This is a solid hypercube. An evil child maliciously replaces some subset of the vector entries by reals of smaller absolute value. Prove or disprove that, no matter what the child does, some nonempty subset of the vectors still will exist that sums to the all-zero vector. (I think I have the answer with a proof, albeit my proof is still a bit sketchy... the definition of "add" can be a bit worrysome when the summands may form some weird, potentially nonmeasureable, set, so your solution will have more juice the more it understands about that.) -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
On 7/2/12, Warren Smith <warren.wds@gmail.com> wrote:
Now consider the set of all vectors in n dimensions, each of whose entries lies in the real closed interval [-1,+1]. This is a solid hypercube. An evil child maliciously replaces some subset of the vector entries by reals of smaller absolute value.
--and I should have said "...and of the same sign, or 0."
You'll need to do something about restricting the values used by the malicious replacer. Otherwise you could take them all linearly independent over Q. Victor On Mon, Jul 2, 2012 at 9:35 PM, Warren Smith <warren.wds@gmail.com> wrote:
On 7/2/12, Warren Smith <warren.wds@gmail.com> wrote:
Now consider the set of all vectors in n dimensions, each of whose entries lies in the real closed interval [-1,+1]. This is a solid hypercube. An evil child maliciously replaces some subset of the vector entries by reals of smaller absolute value.
--and I should have said "...and of the same sign, or 0."
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 7/2/2012 6:27 PM, Warren Smith wrote:
Now consider the set of all vectors in n dimensions, each of whose entries lies in the real closed interval [-1,+1]. This is a solid hypercube. An evil child maliciously replaces some subset of the vector entries by reals of smaller absolute value. Prove or disprove that, no matter what the child does, some nonempty subset of the vectors still will exist that sums to the all-zero vector.
(I think I have the answer with a proof, albeit my proof is still a bit sketchy... the definition of "add" can be a bit worrysome when the summands may form some weird, potentially nonmeasureable, set, so your solution will have more juice the more it understands about that.)
Hmmm. The zero vector is necessarily left (since 0 can't be replaced by a smaller real). So it seems the subset consisting of the zero vector is a trivial solution. Brent Meeker
participants (3)
-
meekerdb -
Victor Miller -
Warren Smith