- #26

mathwonk

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my point is that the technique of infinite cardinality is wasted on the original question, which is easily solved without that notion.

as you have been showing, if it is used, much stronger results can be obtained. my point was the opposite, that if only the weak original result is desired, much easier methods suffice. it seemed worth pointing out since apparently halmos himself did not realize it, according to his final comment as quoted above in post #1.

the uncountability result also seems quite easy. e.g. if B is a countable basis of a vector space V over Q, then elements of V are functions B-->Q which are zero except at a finite set of points of B. Hence they can be partitioned into those in S(n) which are zero except at n pts, for n = 1,..... This is a countable union of the sets S(n), and each S(n) is a union of copies of the countable set Q^n, one for each way of choosing n points from B. Now all ways of choosing n points from B are parametrized by (a subset of) B^n, also countable. thus V is countable, and hence if V is uncountable, it must have uncountable dimension over Q.

Someone has probably done this above, but i could not fully absorb every post quickly.

let me pose another extension of the problem. define an "algebraic number" as one which is a root of a non zero polynomial over Q. then show the algebraic numbers form a vector space of countably infinite dimension over Q, in particular are countable. hence since the reals are uncountable, there must exist transcendental numbers and indeed uncountably many of them.

this is a more substantial consequence of the theory of infinite cardinality.

as you have been showing, if it is used, much stronger results can be obtained. my point was the opposite, that if only the weak original result is desired, much easier methods suffice. it seemed worth pointing out since apparently halmos himself did not realize it, according to his final comment as quoted above in post #1.

the uncountability result also seems quite easy. e.g. if B is a countable basis of a vector space V over Q, then elements of V are functions B-->Q which are zero except at a finite set of points of B. Hence they can be partitioned into those in S(n) which are zero except at n pts, for n = 1,..... This is a countable union of the sets S(n), and each S(n) is a union of copies of the countable set Q^n, one for each way of choosing n points from B. Now all ways of choosing n points from B are parametrized by (a subset of) B^n, also countable. thus V is countable, and hence if V is uncountable, it must have uncountable dimension over Q.

Someone has probably done this above, but i could not fully absorb every post quickly.

let me pose another extension of the problem. define an "algebraic number" as one which is a root of a non zero polynomial over Q. then show the algebraic numbers form a vector space of countably infinite dimension over Q, in particular are countable. hence since the reals are uncountable, there must exist transcendental numbers and indeed uncountably many of them.

this is a more substantial consequence of the theory of infinite cardinality.

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