Tagged Questions

7
votes
1answer
172 views

Pathological linear functionals and ZF

Let $S$ be an infinite set. Let $C(S)$ be the vector space of all functions $S \to \mathbb{R}$, and let $C_c(S)$ be the subspace of functions of finite support. Is the existence of a nonzero linear ...
2
votes
1answer
64 views

Where in this argument ultrafilter is used?

http://en.m.wikipedia.org/wiki/Dimension_theorem#section_1 Let's first not assume any choice principle. Let $V$ be a vector space over a field $F$ and $\beta_1,\beta_2$ be bases for $V$. Suppose ...
6
votes
1answer
126 views

Is axiom of choice required for there to be an infinite linearly independent set in a (non-finite-dimensional) vector space?

In discussing this answer, I noted that while the statement: Any vector space has a basis is equivalent to the axiom of choice, I wondered if the statement that: Any vector space either has ...
-4
votes
1answer
120 views

Proving that a vector space $\mathbb{R}^k, k\in \mathbb N$ has a basis with ZF (and no Axiom of Choice) [duplicate]

Possible Duplicate: Finite dimensional subspaces of a linear space I know that "every vector space has a basis" is equivalent to the "Axiom of Choice". My question: Can I prove that ...
7
votes
3answers
794 views

What is a basis for the vector space of continuous functions?

A natural vector space is the set of continuous functions on $\mathbb{R}$. Is there a nice basis for this vector space? Or is this one of those situations where we're guaranteed a basis by invoking ...
5
votes
1answer
227 views

Does the existence of a $\mathbb{Q}$-basis for $\mathbb{R}$ imply that choice holds up to $\frak c$?

The axiom of choice is, for ZF, equivalent to the statement that every vector space has a basis. The implication of AoC by the existence of a basis for any vector space is shown in this paper. The ...
16
votes
3answers
607 views

Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$?

Assuming the axiom of choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the ...
7
votes
1answer
193 views

Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?

$\mathbb R$ is an uncountably dimensional vector space over $\mathbb Q.$ We can define as many endomorphisms of this vector space as we want by picking their values on the elements of the basis. ...
2
votes
3answers
164 views

Finite dimensional subspaces of a linear space

Suppose $V$ is an infinite dimensional vector space. I do not want to assume the axiom of choice, so I will define a vector space $V$ to be infinite dimensional if there is a proper subspace ...
2
votes
1answer
364 views

When is the pullback of a linear injection a surjection on dual space?

Due to the contravariance of the dual space functor on vector spaces, one might expect the pullback of an injection to be a surjection, and the pullback of a surjection to be an injection. Indeed, for ...
38
votes
2answers
2k views

Axiom of choice and automorphisms of vector spaces

A classical exercise in group theory is "Show that if a group has a trivial automorphism group, then it is of order 1 or 2." I think that the straightforward solution uses that a exponent two group is ...
2
votes
1answer
339 views

Dimension of the sequence space and its dual, depending on status of (AC) and (CH)

Let's consider the sequence space $E =\mathbb R^{\mathbb N}$. If I believe in Choice, I have an isomorphism $E \simeq \mathbb R^{(\mathfrak c)}$ for some cardinal $\mathfrak c$. I further have some ...