Is R a Finite-Dimensional Vector Space Over Q?

[matheinste]

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.

 

[HallsofIvy]

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 ${r_1, r_2, ..., r_n}$. Then every real number would be of the form $a_1r_1+ a_2r_2+ \cdot\cdot\cdot+ a_nr_n}$ where each $a_i$ is a rational number. Then each set of numbers {$a_ir_i$} 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.

 

[matheinste]

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.

 

[andytoh]

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 ${r_1, r_2, ..., r_n}$. Then every real number would be of the form $a_1r_1+ a_2r_2+ \cdot\cdot\cdot+ a_nr_n}$ where each $a_i$ is a rational number. Then each set of numbers {$a_ir_i$} 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).

 

[HallsofIvy]

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!

 

[andytoh]

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.

 

[morphism]

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 = $\cup_n$ F_n. On the other hand, |F_n| <= |Q^n| |B^(<w)| = $\aleph_0$ |I| = |I| (where B^(<w) denotes the finite subsets of B). Whence |R| <= $\aleph_0$|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.)

 

[andytoh]

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|+...= $\aleph_0$ +$\aleph_0$ +$\aleph_0$ +... = $\aleph_0$$\aleph_0$ = $\aleph_0$ ? If so, that is our contradiction.

 

[morphism]

Isn't |R| = |F_1|+|F_2|+|F_3|+... <= |Q|+|Q^2|+|Q^3|= $\aleph_0$ +$\aleph_0$ +$\aleph_0$ +... = $\aleph_0$x$\aleph_0$ = $\aleph_0$ ? 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...

 

[andytoh]

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|...

 

[andytoh]

If |R|<=|I|=|B^(<w)|=|number of subsets of B|=2^|B|=2^$\aleph_0$ = 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.

 

[morphism]

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|.

 

[andytoh]

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|.

 

[morphism]

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.

 

[morphism]

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) $\mapsto$ {b_1, ..., b_n}.
(2) Extend this to F : $\cup_n$ B^n -> C.
(3) |$\cup_n$ 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} $\mapsto$ (b_1, ..., b_n). Deduce that |C_n| <= |B^n| = |B| (well, except when n=0).
(3) |C| = |$\cup_n$ 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...)

 

[andytoh]

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.

 

[morphism]

That's fine; it's more or less the second method I posted in #15.

 

[andytoh]

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?

 

[morphism]

I don't see anything wrong with it!

 

[andytoh]

Thanks morphism. This very interesting topic ends perfectly.

 

[WWGD]

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.

 

[morphism]

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.

 

[mathwonk]

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.

 

[morphism]

Yes, that problem was settled by HallsofIvy in post #2!

 

[andytoh]

We decided to upgrade the original poster's problem to show that the dimension is uncountable as well as infinite. That's all.

 

[mathwonk]

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.

 

[sihag]

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]

you are assuming very difficult transecendence results without proof.

 

[gel]

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?

 

[cogito²]

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".

I had a thread a while back thinking about related questions here: https://www.physicsforums.com/showthread.php?t=177883

 

[sihag]

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]

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.

 

[sihag]

oops, i missed that ! : )

 

[HallsofIvy]

And what could be neater than to quote me?