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Topic: F^I isomorphic to finite(F^I)
Replies: 11   Last Post: Mar 8, 2013 3:45 PM

 Messages: [ Previous | Next ]
 magidin@math.berkeley.edu Posts: 11,665 Registered: 12/4/04
Re: F^I isomorphic to finite(F^I)
Posted: Mar 6, 2013 4:13 PM

On Tuesday, March 5, 2013 5:03:27 PM UTC-6, Kaba wrote:
> Hi,
>
>
>
> Let F be a field, and I be a set. Denote by finite(F^I) the set of
>
> functions from I to F which are non-zero only for finitely many i in I.
>
>
>
> Claim
>
> -----
>
>
>
> F^I is isomorphic (as a vector space over F) to finite(F^I) if and only
>
> if I is finite.

F^I is the dual of finite(F^I); so the statement is equivalent to:

PROPOSITION: A vector space V is isomorphic to its dual if and only if dim(V) is finite.

You can see a collection of proofs at
http://mathoverflow.net/questions/13322/

Alternatively:

From an old post of Bill Dubuque's,

of from

http://math.stackexchange.com/a/58598

we have that if dim(V)=d, and either d or |F| are infinite, then |V|=d|F|=max{d,|F|}.

If d is finite, then isomorphism between V and its dual is classical; in you language, it is clear that F^I and finite(F^I) are the same when I is finite. So we may assume that |I|=d is infinite.

Then V is isomorphic to finite(F^d) and the dual is isomorphic to F^d. The cardinality of finite(F^d) is d|F| = max{d,|F|}.

Now, note that dim(F^d) is at least |F|: if |F| is finite then this is immediate, since V is infinite and embeds into F^d; and if |F| is infinite, fix c=/=0, in F, and define f_c to be element of F^d that has c^n in the nth component for each n in omega, 0 elsewhere (fix an embedding of omega into d). If c_1,...,c_m are pairwise distinct, then f_{c_1},..., f_{c_m} are linearly independent, since

alpha_1f_{c_1} + ... + alpha_mf_{c_m} = 0

implies that for each i we have

alpha_1 c_1^i + ... + alpha_m c_m^i = 0.

Taking the first m of these and vieweing it as a system of m equation in the m unknowns alpha_1,...,alpha_m, the coefficient matrix is a Vandermonde matrix, and since the c_i are pairwise distinct, the determinant is nonzero. Thus, the only solution is alpha_1=...=alpha_m = 0. Therefore, F^d contains at least |F| linearly independent vectors, so dim(F^d)>=|F|.

Let k = dim(F^d). Then |F^d| = k|F| = max(k,|F|). But k = dim(F^d)>=|F|, so |F^d| = k.

Hence

dim(F^d) = k = |F|^d >= 2^d > d = dim(finite(F^d)). Therefore, the two are not isomorphic, since they don't have the same dimension.

> In addition to the proof, some interesting questions arise:
>
>
>
> 1) What is the cardinality of F^I?

Assuming the Axiom of Choice:

For finite |I|, it is |F|^|I|; if F is finite that's the best you can say; and if F is infinite then this is |F|.

If |I| is infinite, and |F|<= 2^|I|, then the cardinality is 2^|I|. We have:

2^|I| <= |F|^|I| <= (2^|I|)^|I| = 2^{|I x I|} = 2^|I|.

If |F|>2^|I|, then it is harder to state exactly what it is; if you assume the Generalized Continuum Hypothesis, then it's |F| if |I|<cofinality(|F|), and the successor of |F| if |I|>= cofinality(|F|).

> 2) What could be a basis for F^I?

You need the Axiom of Choice to guarantee the existence of a basis; in the absence of the Axiom of Choice, it is consistent with ZF that the direct product of countably many copies of the field of 2 elements does not have a basis. So I don't think you can *exhibit* a basis, or even generally describe it.

> 3) What is the dimension of F^I?

See above; that's the best you can say in the absence of AC and the Generalized Continuum Hypothesis.

--
Arturo Magidin

Date Subject Author
3/5/13 Kaba
3/5/13 Shmuel (Seymour J.) Metz
3/6/13 Kaba
3/6/13 David C. Ullrich
3/8/13 Kaba
3/8/13 Kaba
3/8/13 David C. Ullrich
3/6/13 Shmuel (Seymour J.) Metz
3/6/13 Robin Chapman
3/6/13 Kaba
3/6/13 magidin@math.berkeley.edu
3/7/13 Kaba