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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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    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.2012-07-26
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    Every vector space has a Hamel basis (which is infinite if $\dim V$ is infinite). What definition of “basis” do you use?2012-07-26
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    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?2012-07-26
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    @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.2012-07-26
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    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.2012-07-26
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    @William: I think your contention is a little strong. ZFC is basically the standard object of study in set and model theory. I'm not saying there are not dissenters, but rejecting ZFC suggests either a constructivist or finitist bent, and I don't think these are mainstream positions among professional mathematicians (though perhaps for complexity theorists).2012-07-26
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    @trb456 Most mathematician would agree to the axion of infinity or that the union of two sets is a set. How many mathematicians have indicated the use of the axiom foundation in anything they do? Rejecting foundation and the axiom of choice hardly makes a person a contructivist or finitist.2012-07-26
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    @William: I'll give you Foundations--Jech (2000) even says this, pointing out it's rather needed for set theory and models but not much else. But rejecting Choice does kinda make you a constructivist, and since it's AC's interaction with infinity that produces the pathologies, it kinda makes you a finitist, too! ;-)2012-07-28
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    @trb456 Rejecting the axiom of choice does not make you a finitist. Alot a set theory is done without it. In fact by a theorem of Hartogs, $ZF$ can prove that there exists uncountable cardinals even without the axiom of choice or the well-ordering principal. That is pretty far from being a finitist.2012-07-28
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    @William: Right, but the pathologies of AC ARE still related to infinity as I said, and without AC there's a whole lot you can't DO with large cardinals, as you clearly must know given your answer. As Munkres noted in his book Topology, the first uncountable ordinal exists without AC, but you can't DO anything with it!2012-07-28
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    @trb456 I don't remember if Munkres ever state this result, but you should be sure you understand the significance of Hartogs Theorem. If $\omega$ exists, you can prove that $\mathscr{P}(\omega)$ is uncountable (not in bijection with $\omega$). However, without choice, this alone does not imply that there is an uncountable ordinal. An ordinal is a very special type of set. Nevertheless, the theorem states there actually does exists an uncountable ordinal.2012-07-28
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    @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

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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?2012-07-26
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    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)?2012-07-26
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    @DonAntonio I think your comment has to do with the typo brought up by trb456.2012-07-26
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    Indeed @William , thanks.2012-07-26