Vector space over the rationals

  • Thread starter matheinste
  • Start date
  • #1
1,060
0
Hello all.

I came across this problem in Halmos, Finite-Dimensional Vector Spaces, page 16.

Is the set R of all real numbers a finite-dimensional vector space over the field Q of all rational numbers. There is a reference to a previous example which says that with the usual rules of addition and multiplication by a rational R becomes a rational vector space. My answer to the question would be that R is not a finite-dimensional vector space over the field Q.

The author goes on to say that the question is not trivial and it helps to know something about cardinal numbers.

Can anyone please expand on this.

Thanks Matheinste.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
964
Hello all.

I came across this problem in Halmos, Finite-Dimensional Vector Spaces, page 16.

Is the set R of all real numbers a finite-dimensional vector space over the field Q of all rational numbers. There is a reference to a previous example which says that with the usual rules of addition and multiplication by a rational R becomes a rational vector space. My answer to the question would be that R is not a finite-dimensional vector space over the field Q.

The author goes on to say that the question is not trivial and it helps to know something about cardinal numbers.

Can anyone please expand on this.

Thanks Matheinste.
You are correct that the set of all real numbers, as a vector space over the rational numbers, is NOT finite-dimensional.

If it were finite dimensional, then there finite basis, say [itex]{r_1, r_2, ..., r_n}[/itex]. Then every real number would be of the form [itex]a_1r_1+ a_2r_2+ \cdot\cdot\cdot+ a_nr_n}[/itex] where each [itex]a_i[/itex] is a rational number. Then each set of numbers {[itex]a_ir_i[/itex]} would be countable because the set of rational numbers is countable. The set of all real numbers would then be a Cartesian product of countable sets. That would imply that the set of all real numbers is countable- but it isn't.
 
  • #3
1,060
0
Thanyyou HallsofIvy.

I know just enough to follow your argument but would not have reasoned it out for myself. That completely answers my query.

Thanks again. Mateinste.
 
  • #4
359
3
You are correct that the set of all real numbers, as a vector space over the rational numbers, is NOT finite-dimensional.

If it were finite dimensional, then there finite basis, say [itex]{r_1, r_2, ..., r_n}[/itex]. Then every real number would be of the form [itex]a_1r_1+ a_2r_2+ \cdot\cdot\cdot+ a_nr_n}[/itex] where each [itex]a_i[/itex] is a rational number. Then each set of numbers {[itex]a_ir_i[/itex]} would be countable because the set of rational numbers is countable. The set of all real numbers would then be a Cartesian product of countable sets. That would imply that the set of all real numbers is countable- but it isn't.

I don't think that set of all real numbers, as a vector space over the rational numbers, is even countable-dimensional, because (using the exact same argument as HallsovIvy), R would then be a countable cartesian product of countables sets, which is not necessarily countable (only a finite cartesian product of countable sets is countable).
 
Last edited:
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
964
Yes, that's right- the dimension of the real numbers, as a vector space over the rational numbers, is not countable. However, the original question just asked about the proof that it was not finite dimensional!
 
  • #6
359
3
Is there a quick proof to why the vector space of reals over the rationals has uncountable dimension? A countable cartesian product of countable sets is not necessarily countable, but it is not necessarily uncountable either.

All that is needed is to construct one such (Hamel) basis, show that it is an uncountable basis. Then all other bases would have the same cardinality and hence be uncountable as well.
 
Last edited:
  • #7
morphism
Science Advisor
Homework Helper
2,015
4
If B={r_i} (i in some infinite index set I) is a basis for R/Q, then each real number can be written as a finite linear combination of the r_i's over Q. Let F_n be the set of real numbers expressible as a linear combination of n elements of B. Then R = [itex]\cup_n[/itex] F_n. On the other hand, |F_n| <= |Q^n| |B^(<w)| = [itex]\aleph_0[/itex] |I| = |I| (where B^(<w) denotes the finite subsets of B). Whence |R| <= [itex]\aleph_0[/itex]|I| = |I|.

I think this is alright. (Although to be completely rigorous, I think we need to employ the well-ordering theorem on B. Try to see where this is needed. Edit: On second thought, I suppose this can be avoided if we assume for a contradiction that I is countably infinite - this way we can give B the induced well-ordering from N.)
 
Last edited:
  • #8
359
3
I didn't know what your I stood for.

Using your notation, don't we simply have |R| = |F_1|+|F_2|+|F_3|+... = |Q|+|Q^2|+|Q^3|+...= [itex]\aleph_0[/itex] +[itex]\aleph_0[/itex] +[itex]\aleph_0[/itex] +... = [itex]\aleph_0[/itex][itex]\aleph_0[/itex] = [itex]\aleph_0[/itex] ? If so, that is our contradiction.
 
Last edited:
  • #9
morphism
Science Advisor
Homework Helper
2,015
4
Isn't |R| = |F_1|+|F_2|+|F_3|+... <= |Q|+|Q^2|+|Q^3|= [itex]\aleph_0[/itex] +[itex]\aleph_0[/itex] +[itex]\aleph_0[/itex] +... = [itex]\aleph_0[/itex]x[itex]\aleph_0[/itex] = [itex]\aleph_0[/itex] ? If so, that is our contradiction.
No. For example F_1 contains a copy of Q for each r_i in B, and F_2 contains a copy of Q for each pair {r_i, r_j} in B. So if I is uncountable (which I am assuming can happen, modulo my remark towards the end), then |F_n| need not be bounded by |Q^n|.

Also, strictly speaking, R isn't a disjoint union of the F_n's because I didn't specify that the n elements taken from B be distinct. But this is immaterial...
 
  • #10
359
3
Let F_n be the set of real numbers expressible as a linear combination of n elements of B, with none of the rational coefficients being zero. By the uniqueness of an element expressed as a linear combinations of basis elements, then we have R = |{0}|+|F_1|+|F_2|+..., since the F_n are now disjoint. I'll look into the bounds of the |F_n|....
 
Last edited:
  • #11
359
3
If |R|<=|I|=|B^(<w)|=|number of subsets of B|=2^|B|=2^[itex]\aleph_0[/itex] = c = |R|,
where is the contradiction? What if we use Schauder bases (allowing for infinite sums of the basis elements)?

There is no injection P(B)-> B, so P(B) is uncountable since B is equivalent to the natural numbers by assumption.
I believe there is no injection from the set of all finite subsets to B either, so B^(<w) would have cardinality >= c.
 
Last edited:
  • #12
morphism
Science Advisor
Homework Helper
2,015
4
Can you stop and read my post from the beginning? It seems like you're completely missing the point.

(1) I'm taking any Hamel basis B whose cardinality is |I|. All we know about |I| is that it's infinite (although, as I indicated in the end of my post, we can assume that I is countable and get a contradiction - by going through the argument unchanged: we get |R| <= |I|).
(2) B^(<w) is the set of finite subsets of B, and not the power set of B. It's an easy exercise to prove that if |B| is infinite, then |B^(<w)|=|B|.
(3) In regards to what you posted in post #10, we don't really need to write R as a disjoint union of the F_n's. It's perfectly sufficient that |R| <= |F_1| + |F_2| + ..., since I already gave you an upper bound for each |F_n|.
 
  • #13
359
3
It's an easy exercise to prove that if |B| is infinite, then |B^(<w)|=|B|.
I want to believe you, but I don't see it (yet).

B^(<w) = the finite subsets of B

Let g: B -> B^(<w). Claim: g cannot be surjective.
Let K={b in B| b does not belong to g(b)}. If g is surjective, let g(x) = K. Then x belongs to K iff x does not belong to g(x)=K, a contradiction.


Oops, K can be infinite. Ok, I'll try to prove that |B^(<w)|=|B|.
 
Last edited:
  • #14
morphism
Science Advisor
Homework Helper
2,015
4
As for Schauder bases, well, I'm only familiar with this concept in the scope of Banach spaces. But I looked it up, and Wikipedia says that a Schauder basis is countable by definition.
 
  • #15
morphism
Science Advisor
Homework Helper
2,015
4
I want to believe you, but I don't see it (yet).

C = the finite subsets of B (the B^(<w))

Let g: B -> C. Claim: g cannot be surjective.
Let K={b in B| b does not belong to g(b)}. If g is surjective, let g(x) = K. Then x belongs to K iff k does not belong to g(k)=K, a contradiction.

Thus there is no injection from C to B.
Why is K finite?

Here's a sketch you can use to prove |C| <= |B|:
(1) For each n, define f_n : B^n -> C by (b_1, ..., b_n) [itex]\mapsto[/itex] {b_1, ..., b_n}.
(2) Extend this to F : [itex]\cup_n[/itex] B^n -> C.
(3) |[itex]\cup_n[/itex] B^n| = |B|.
(4) Try to reason that |C| <= |B|.

Alternative path:
(1) Let C_n = { A in C : |A| = n }.
(2) Well-order B. Define f : C_n -> B^n by {b_1 < ... < b_n} [itex]\mapsto[/itex] (b_1, ..., b_n). Deduce that |C_n| <= |B^n| = |B| (well, except when n=0).
(3) |C| = |[itex]\cup_n[/itex] C_n| <= |B|.
(I essentially used these ideas in post #7. First I decided not to reuse them here, but then I figured I might as well...)
 
Last edited:
  • #16
359
3
I didn't read your proof to why |C|=|B| yet, but while I slept I thought of the following proof:

Let B = {b_1,b_2,...}. Let B_k be the collection of all subsets of B whose element with the highest index is b_k. Then the elements of B_k is mapped bijectively to any set with 2^(k-1) elements (the number of subsets of {b_1,...,b_(k-1)}. Then C = U(B_k) is mapped bijectively to a countable collection of finite sets and hence is countable, and so |C|=|B|.

In my proof, I assumed that B is countable. If B is not countable, I suppose one can just well-order the index of B and use transfinite induction.
 
Last edited:
  • #17
morphism
Science Advisor
Homework Helper
2,015
4
That's fine; it's more or less the second method I posted in #15.
 
  • #18
359
3
I finished my proof that |C|=|B| for the uncountable case (using your method, because your f in your second method is injective). I was wondering if a proof using transfinite induction could work too. Not enough exercises in transfinite induction are given in textbooks.

For readers not wishing to read the previous posts: B is any uncountable set and C is the collection of all finite subsets of B. Use transfinite induction to prove that |C|=|B|.

Call I the index set for B and well-order I. Let J be all the elements of I such that the collection C(J) of all finite subsets of B with elements indexed by J has cardinality <= B. If the section S_i is a subset of J, then the only new finite subsets created by introducing {i} are just KU{b_i}, where K belongs to C(J). So then
|C(JU{i})| <= |C(J)| + |C(J)| = |C(J)| <= |B|, so that {i} belongs to J. Thus J is inductive so that J = I by the principle of transfinite induction.

Is that right?
 
Last edited:
  • #19
morphism
Science Advisor
Homework Helper
2,015
4
I don't see anything wrong with it!
 
  • #20
359
3
Thanks morphism. This very interesting topic ends perfectly.
 
  • #21
WWGD
Science Advisor
Gold Member
5,529
4,229
Hope this is not too simple of a question. I have not worked with infinite-dim.

V.Spaces, just read a bit in Wikipedia . The basis is infinite,(until dust settles and

it is clear if it is countable. Maybe we could look at L.Indep. elements and their

maximal cardinality)

or not) but, given that it is an algebraic basis, are

linear combinations finite, i.e, finite selections of an infinite set?.


I mean, given the basis B , and x in IR , do we express:

x=Sum(i=1,..,n) q_ib_i

for b_i in B , q_i in Q ?.

Or do we use infinite sums -- but then we need to worry about convergence.

I think in this last case, if we worried about convergence, so I imagine we then need

a norm, or topology to talk about convergence, then B would be a

Schauder (Chowdah? ) basis, right?

Thanks.
 
  • #22
morphism
Science Advisor
Homework Helper
2,015
4
In most cases (namely when one talks about Hamel bases, and this is almost all the time) we only take finite sums. And yes, Schauder bases are mentioned when we have a topology, almost always normed, and here the sums may be infinite.
 
  • #23
mathwonk
Science Advisor
Homework Helper
2020 Award
11,154
1,349
this problem has nothing to do necessarily with uncountably infinite cardinals.

for every prime p, the polynomial X^p-1 has an irreducible factor of degree p-1 over Q.

the root field of this polynomial has dimension p-1 over Q, hence the complex numbers have dimension >p-1 over Q for all primes p, hence they have infinite dimension over Q.

since the complexes have dimension 2 over R, R has infinite dimension over Q.

so this problem is in fact almost trivial, and needs nothing about uncountability. hence apparently halmos is wrong here to overestimate its difficulty.
 
Last edited:
  • #24
morphism
Science Advisor
Homework Helper
2,015
4
Yes, that problem was settled by HallsofIvy in post #2!
 
  • #25
359
3
We decided to upgrade the original poster's problem to show that the dimension is uncountable as well as infinite. That's all.
 

Related Threads on Vector space over the rationals

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
10
Views
4K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
4
Views
732
Replies
4
Views
3K
Replies
5
Views
2K
Replies
1
Views
2K
Replies
4
Views
6K
Top