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Would you tell me why the statement below holds?

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

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10  
The statement is blatantly false: a vector space, any vector space, always has a basis (as long as we accept the usual mathematical logical ZF + AC). Read/check/think carefully what you actually want to ask. –  DonAntonio Jul 26 '12 at 1:55
    
Every vector space has a Hamel basis (which is infinite if $\dim V$ is infinite). What definition of “basis” do you use? –  Yury Jul 26 '12 at 1:58
    
I recall when taking linear algebra back in the day encountering a question like this. We never handled infinite dimensional spaces, so it seemed to me like a poor question to ask without introduction. The definition of a basis for an infinite dimensional space does not follow easily from the finite-dimensional equivalent. Is this a question from Axler, perhaps? –  Arkamis Jul 26 '12 at 2:00
2  
@DonAntonio is exactly right. "Every vector space has a basis" is equivalent to the Axiom of Choice (AC), so unless we are rejecting AC, the question is false. –  trb456 Jul 26 '12 at 2:01
    
I wonder how many mathematician would say that ZFC is the axiom of mathematics. Many set theorist would not agree. And frankly, if you told mathematicians the statement of the axiom of foundation and the axiom of choice, how many would agree. –  William Jul 26 '12 at 2:17

2 Answers 2

up vote 10 down vote accepted

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.

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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.

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"Cannot" instead of "can" prove in the last sentence? –  trb456 Jul 26 '12 at 2:14
    
About your last statement , @William: (1) How AC being independent of the other axioms in ZF makes it possible to prove within ZF but without AC that every v.s. has a basis? (2) How can be proved actually that we can construct a maximal lin. independent set of vectors for any v.s. without resourcing to AC or something equivalent to it in ZF (say, Zorn's Lemma)? –  DonAntonio Jul 26 '12 at 2:16
    
@DonAntonio I think your comment has to do with the typo brought up by trb456. –  William Jul 26 '12 at 2:19
    
Indeed @William , thanks. –  DonAntonio Jul 26 '12 at 2:20

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