1
$\begingroup$

Would you tell me why the statement below holds?

A vector space $V$ has a basis if and only if $0 < \dim V < \infty.$

  • 0
    @William: I do get this (Jech (2000) goes over what we know and don't know without AC quite thoroughly). What Munkres is saying is that we can't DO much with the knowledge of uncountable sets without well-ordering them. And this comports with finitists quite well: they don't deny infinite objects, they deny we can do anything with them. It's the ULTRAfinitists that deny that infinite objects exists.2012-07-28

2 Answers 2

11

The claim is false, whether or not one accepts the Axiom of Choice.

Consider the vector space $V$ (over the reals) of all polynomials $P(x)$ with real coefficients. Addition of polynomials, and multiplication by a scalar, are defined in the usual way.

The space $V$ is infinite-dimensional, but has basis $\{1,x,x^2,x^3,x^4,\dots, x^n, \dots\}$.

Remark: It is not difficult to show, without using the Axiom of Choice, that any finite dimensional space does have a basis, so the implication in one direction is true.

1

Your claim is not true. There certainly exists infinite dimensional vector spaces.

However, the statement that every vector space has a basis is provable depending on your axioms.

For example, $ZF$ without foundation and without the power set axiom but with the well-ordering principle can prove that every vector space has a basis.

Moreover, ZF with the fact that every vector space has a basis can prove the axiom of choice (and hence the well-ordering principle). See Existence of Bases Implies the Axiom of Choice by Blass. This show that over $ZF$, the axiom of choice is equivalent to the fact that every vector space has a basis.

Also the axiom of choice is independent of $ZF$, so using just $ZF$ alone you cannot prove that every vector space has a basis.

  • 0
    Indeed @William , thanks.2012-07-26