
Re: F^I isomorphic to finite(F^I)
Posted:
Mar 6, 2013 4:13 PM


On Tuesday, March 5, 2013 5:03:27 PM UTC6, 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 nonzero 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,
https://groups.google.com/forum/?fromgroups=#!msg/sci.math/8aeaiKMLP8o/OeF_n4IseYIJ
of from
http://math.stackexchange.com/a/58598
we have that if dim(V)=d, and either d or F are infinite, then V=dF=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 dF = 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 = kF = 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

