Re: Example of a Vector Space which cannot be made into an inner product space
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 6 Dec 2005 16:07:04 +0000 (UTC)
In article <un3bp1d2q2tlr59rp2pfmf2u3bq07pl4i1@xxxxxxx>,
David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
>On 5 Dec 2005 12:02:27 -0800, "Jules" <julianrosen@xxxxxxxxx> wrote:
>
>>
>>[...]I believe the Axiom of
>>Choice implies that every vector space has a basis, in the sense that
>>any vector can be written as a finite linear combination of vectors in
>>the basis (I not positive about this, though).
>
>This is easy: Zorn's Lemma implies that there is a maximal
>linearly independent set, and maximality implies it must
>be a basis. (The converse mentioned by Arturo, that
>"every vector space has a basis" implies AC, is not
>so trivial, I think.)
It is not, indeed.
I can give a slightly weaker statement, with a proof due to Halpern,
simpliflied by Lauchli.
Consider the statement
(SB) If V is any vector space, and S is a spanning set for V, then
there exists a basis B of V which is contained in S.
THEOREM. In ZF, (SB) -> AC
Proof. Let Y = {A_i} i in I be a nonempty family of pairwise disjoint
nonempty sets. Let F be a field of cardinality no less than
U_{i in I}A_i = U(Y), and fix a one-to-one map U(Y)->F-{0}. This can
be done without invoking AC; for example, take Q(U(Y)), the fraction
field of the ring of multivariable polynomials with coefficients in Q,
and where the variables are the elements of the A_i; the obvious
embedding will work then.
Denote the image of x in U(Y) by x*.
Consider the vector space F^Y of all functions from Y to F, with
pointwise addition and scalar multiplication. Let V be the subspace of
F^Y consisting of the functions of finite support:
V = { f in F^Y : f(A_i) = 0 for all but finitely many i in I}.
Given an element x in U(Y), define the function (x):Y->F by
(x)(A_i) = x* if x in A_i
(x)(A_i) = 0 otherwise.
The functions are elements of V, since the A_i are pairwise disjoint.
The claim is that the set {(x) : x in U(Y)} spans V. If f is in V,
then f(A_i)=0 for almost all i. If i_1,...,i_n in I are such that
f(A_{i_j}) is nonzero, then let x_j in A_{i_j} for j=1,...,n (this can
be done without invoking AC because there are only finitely many
choices involved), and let b_j = f(A_{i_j}). Then
f = b_1(x_1) + ... + b_n(x_n)
proving that the set spans V. By (SB), there is a basis B contained in
the set {(x) : x in U(Y)}. Let C be the set
C = { x in U_{i in I} A_i : (x) in B }.
The claim is that C/\A_i is a singleton for each i.
First, assume that x,y are in C/\A_i for some i. If x<>y, then
y*(x)=x*(y), because both have value 0 at A_j if j<>i, and value y*x*
at A_i. Since x<>y it follows that x*<>y*, and therefore {(x),(y)} is
linearly dependent, contradicting the fact that B is a basis. So
C/\A_i has at most one element for each i.
Next, if C/\A_i is empty for some i, then all elements (y) in B
satisfy (y)(A_i) = 0; but then, letting x in A_i, we have that
(x)(A_i)=x*<>0, so (x) cannot be a linear combination of the elements
of B, contradicting the fact that B spans V. Therefore, C/\A_i has at
least one element for each i.
Thus, C/\A_i is a singleton for each i. Define f:I->U(Y) by letting
f(i) be the only element of C/\A_i. This is a choice function, thus
establishing AC. QED
The statement that "Every vector space has a basis" also implies AC;
the proof given by Andreas Blass is to show that this statement
implies the Axiom of Multiple Choice (in a slightly weaker axiomatic
scheme than ZF). Multiple Choice states that given a nonempty family
{A_i} of pairwise disjoint sets, there is a function f that assigns to
each i in I a finite subset of the A_i; the proof runs along similar
lines to Halpern's proof above. And then Blass referes to Jech's book
"The axiom of choice" (to which I do not have access) to show that in
ZF, Multiple Choice implies Choice; so in fact one can show that
"Every vector space has a basis" implies AC directly, though the proof
does seem to be more complicated.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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