2
votes
2answers
65 views

Can we prove without the axiom of choice that subspaces of finite dimensional vector spaces are finite dimensional as well?

Let $E$ be a vector space and $F$ a subspace of $E$. Show that if $E$ is finite dimensional then $F$ is finite dimensional too. It's easy to prove by contradiction by taking a family linearly ...
1
vote
0answers
38 views

Hilbert space without the projection theorem

One succinct statement of the projection theorem in Hilbert space is $A+A^\bot=\scr H$, where $A\in\scr C$, the set of closed subspaces of $\scr H$. (We will also denote the set of all subspaces by ...
1
vote
0answers
61 views

Hilbert projection theorem without countable choice

All the proofs of the Hilbert projection theorem, existence part, that I have seen so far use countable choice (usually implicitly). Is this necessary? It seems like you might be able to leverage the ...
0
votes
2answers
50 views

Is this a basis for the dual space?

There is an example on Wikipedia that I don't understand and I'd appreciate some help. They define $\mathbb R^\infty$ to be the space of all sequences that are zero except for finitely many indexes. ...
0
votes
1answer
139 views

Infinite dimensional vector space. Linearly independent subsets and spanning subsets

This question is a follow up to this question: The dimension of the real continuous functions as a vector space over $\mathbb{R}$ is not countable? I realized that the answer I accepted used an ...
2
votes
0answers
185 views

Hahn-Banach via Hamel Basis

my question for tonight: Is there a proof for Hahn-Banach using a Hamel Basis? I know, the proof for existence of Hamel Bases uses already Axiom of Choice, but I'd like to apply this without refering ...
1
vote
1answer
91 views

Axiom of choice and vector space bases

Is this: for every vector space $V$, if $B$ and $C$ are bases of $V$, then there is a bijection: $B\to C$ iff the axiom of choice holds true? Or, perhaps, if axiom of choice is replaced by ...
2
votes
1answer
250 views

Algebraic complements in vector space of functions without the axiom of choice

The axiom of choice is equivalent to the statement that every subspace $U$ of every vector space $V$ has an algebraic complement, i.e. another subspace $W$ that has a trivial intersection with the ...
2
votes
3answers
696 views

Prove that Every Vector Space Has a Basis

My textbook extended the following proof to show that every vector space, including the infinite-dimensional case, has a basis. Condition: $S$ is a linearly independent subset of a vector space ...
7
votes
2answers
158 views

Vector space bases without axiom of choice

I want to find an example of a vector space with no base if we assume that axiom of choice is incorrect. This question might be duplicate so please alert me. Thanks.
7
votes
1answer
210 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
72 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 ...
7
votes
1answer
211 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
144 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 ...
9
votes
3answers
2k 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
332 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 ...
18
votes
3answers
729 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
252 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
242 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
586 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 ...
3
votes
1answer
433 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 ...