# Vector space over the rationals

mathwonk
Homework Helper
2020 Award
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.

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R is an extension field of Q
To show [R:Q] is infinite.
Of course R is a vector space over Q.
There's a result which states an extension field is algebraic if and only if it is the union of finite subextensions.
'pi' and e are transcendental elements of R over Q.
So, R is not algebraic and hence comprises of an infinite subextension of Q.
It follows that [R:Q] is infinite.
Proved, right ?
Or have I missed something?

mathwonk
Homework Helper
2020 Award
you are assuming very difficult transecendence results without proof.

I have a further question about the vector space R/Q or C/Q (the complex numbers as a vector space over the rationals).
Given any transcendental number a, then $\{a^n:n\in Z\}$ is a countably infinite linearly independent set. Can anyone explicitly construct an uncountable LI set?

Also, I once worked out that $\{\exp(x):x {\rm\ is\ algebraic}\}$ is LI over the rationals (in fact, it is LI over the algebraic numbers). This is still countable, but it is quite interesting. Are there any other such interesting LI sets?

Given any transcendental number a, then $\{a^n:n\in Z\}$ is a countably infinite linearly independent set. Can anyone explicitly construct an uncountable LI set?

Another similar question might be: Can anyone come up with an explicit Hamel basis of an infinite dimensional (separable) Hilbert space? I don't think it's possible (in most meanings of "explicit"). I mean basically the reason you throw analysis into separable infinite dimensional Hilbert spaces is so that you can take a space with an uncountable (Hamel) basis and instead alter your meaning of "basis" to one which is countable (which will work fine when doing stuff with continuous operators). Then everything is countable and thus more naturally "constructible".

something much neater!

If a vector space V is finite dimensional , of dim n say, over a field F
then, (assume, {v1,v2,.....,vn} is a basis for V)
V is isomorphic to F^n
through the correspondence
a1.v1 +a2v2 + a3v3 + ..... + anvn -----> (a1,a2,a3,.....,an) {the n component tuple of F^n) (verify the bijective linear transformation)
It follows that R (reals) would be isomorphic to Q^n (Q denotes field of rationals)
But, Q^n is countable (Q being countable)
implying that R is countable (absurd)
Q.E.D.

morphism
Homework Helper
something much neater!

If a vector space V is finite dimensional , of dim n say, over a field F
then, (assume, {v1,v2,.....,vn} is a basis for V)
V is isomorphic to F^n
through the correspondence
a1.v1 +a2v2 + a3v3 + ..... + anvn -----> (a1,a2,a3,.....,an) {the n component tuple of F^n) (verify the bijective linear transformation)
It follows that R (reals) would be isomorphic to Q^n (Q denotes field of rationals)
But, Q^n is countable (Q being countable)
implying that R is countable (absurd)
Q.E.D.
How is this neater? It's exactly what HallsofIvy had in post #2.

oops, i missed that ! : )

HallsofIvy