Independence of axioms

Question

What does it really mean for a set of axioms to be independent? In particular, say there are two persons $A$ and $B$, who assume and do not assume the axiom of choice respectively. So, $A$ can prove that every vector space has a basis. However, if $B$ takes an arbitrary vector space and gives it to $A$, then $A$ can produce a basis. So can't $B$ just use the basis that $A$ just produced? But then, every vector space would have a basis even if the Axiom of Choice were not assumed.

Answer 1

For a set of axioms $T$ to be independent it means that if $\varphi\in T$, then $\varphi$ is not a logical consequence of $T\setminus\{\varphi\}$.

In the case of the axiom of choice, and the rest of set theory, $\sf ZF$, you're mistaken. If we are in a universe without the axiom of choice, we cannot even ensure that $\Bbb R$ has a basis as a vector space over $\Bbb Q$. You are thinking about "passing the reals to another person", but you're invariably changing the universe in which you are working when you assume choice.

Even worse, set theory has taught us that given any set, it can be a countable set in a slightly larger universe of mathematics. And indeed every countable vector space has a basis. So given any vector space, we can always go to a larger universe where it has a basis. This is really not the point. The point is that in a given universe of mathematics, a vector space may or may not have a basis. Assuming different background theory means that you might be working in a different universe with more—or less—sets.

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Here is a simpler example.

What does it mean that $\sqrt2$ "exists"? You are thinking about "all the mathematical universe", but if you are just given an ordered field $F$, you don't know a priori that $2$ has a square in $F$. It might be the rational numbers, where this is false and it might be the real numbers where this is true.

To make matters worse, we can still talk about $\sqrt2$ even if it doesn't exist, as it is the unique positive solution for $x^2-2=0$.

So does $\sqrt2$ exist? Again, we come to the problem of "what is mathematical existence?", and the answer is that existence if relative to some fixed universe of interest. In $\Bbb Q$, $\sqrt2$ does not exist; but in $\Bbb R$ it does exist.

Going back to set theory, even if you have the same vector space in two different universes of set theory, its subsets might be different. In particular, it might be that in one universe it has a basis and in another it does not. (And to add insult to injury, there is also the matter of confusing the object with its definition, which is a whole other story and makes matters far more difficult to grasp intuitively.)

Answer 2

Your basic mistake is here:

However, if B takes an arbitrary vector space and gives it to A, then A can produce a basis.

Using the axiom of choice, A can prove that a basis exists, but he cannot construct that basis in order to give it to B. Since B doesn't assume the axiom of choice, he'll not accept A's proof which relies on that axiom, and therefore still can maintain that there might be a vector space without a basis.

B can't prove that such a vector space exists, of course, unless he assumes another additional axiom that allows him to prove it; that other axiom would then of course be incompatible with the axiom of choice (and certainly would be rejected by A).

To make a concrete example, it is easy to show that the set of real numbers is a vector space over the set of rational numbers. Note that both the set of rational numbers and the set of real numbers can be constructed from the ZF axioms without using the axiom of choice; A and B can agree on a common construction (e.g. the construction of the real numbers via Dedekind cuts on the rational numbers, the successive construction of the rational numbers from the natural numbers via equivalence classes of pairs, the Kuratowski construction of pairs, and the construction of natural numbers as finite von-Neumann ordinals).

Now using the axiom of choice, A can prove using the axiom of choice that this vector space has a basis. However it cannot be proven without the axiom of choice, so B will not be able to prove that. Now your suggestion is that A just produces a basis, gives it to B, and then B can verify that it is a basis. However, A is not able to produce a basis. There is no way to write down a construction that specifies a specific basis of $\mathbb R$ as a vector space over $\mathbb Q$. What A can do is write down a construction which at some time contains “now you can choose …”, but this invokes the axiom of choice, and when A gives that to B, B will respond with “Sorry, I don't see how I can choose. Please tell me how to construct a choice function for this case.” But A cannot; all A can do is to say that because of the axiom of choice there must exist one. But B doesn't have the axiom of choice, and therefore cannot accept that argument.

Answer 3

I have the impression the other answers don't go to the root of the misunderstanding so I will try to contribute this viewpoint. The real problem here is realist views concerning the existence of mathematical objects. If sets and vector spaces are really out there, then once person A produces a basis for a vector space, what's to stop person B from using it? To understand what is going on you have to step back and realize that such presumed existence is a fairy tale fed by beliefs in the so-called intended interpretation. Once viewed from such a distance the OP may be able to follow the remarks concerning the dependence of properties of sets on the universe one is in.