I About the existence of Hamel basis for vector spaces

[cianfa72]

TL;DR Summary

Does an Hamel basis always exist in any specific instance of a vector space?

It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).

Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?

 

[Gavran]

It can be shown that if the Axiom of Choice is false, then there are vector spaces with no Hamel basis. The exclusion of the Axiom of Choice means the exclusion of the proof based on the Axiom of Choice and this does not mean there are vector spaces with no Hamel basis. Every vector space has a Hamel basis.

 

[cianfa72]

It can be shown that if the Axiom of Choice is false, then there are vector spaces with no Hamel basis. The exclusion of the Axiom of Choice means the exclusion of the proof based on the Axiom of Choice and this does not mean there are vector spaces with no Hamel basis. Every vector space has a Hamel basis.

Ah ok. So your claim is that every vector has an Hamel basis even though it can't proved without assuming the Axiom of Choice (AC). Basically the exclusion of the AC rules out just the proof based on it but not the conclusion (i.e. every vector space has an Hamel basis).

 

[Gavran]

Ah ok. So your claim is that every vector has an Hamel basis even though it can't proved without assuming the Axiom of Choice (AC). Basically the exclusion of the AC rules out just the proof based on it but not the conclusion (i.e. every vector space has an Hamel basis).

A ⇔ B
The exclusion of A does not produce B is false.

 

[nuuskur]

It is known that the axiom of choice (in its full power) is equivalent to the statement that any vector space has a basis. For finite dimensional spaces, we don't need choice to be able to produce a basis. Without choice, there is no hope for a basis in $\mathbb R_\mathbb Q$ for example. Even moreso, without choice, there is no reasonable way to define dimension for an arbitrary vector space (e.g when there are bases* of different cardinalities). One could say the space is infinite dimensional, if it cannot be generated by finitely many elements and leave it at that.

* by a basis I mean a minimal generating subset here

Also strictly speaking, I don't know if not(AC) is sufficient to guarantee a vector space without basis, maybe something stronger is required. But $\mathbb R_\mathbb Q$ is an infamous example of one where you can't get a basis without choice, but that doesn't yet imply it has no basis.

 

[cianfa72]

A ⇔ B
The exclusion of A does not produce B is false.

The above $A \Leftrightarrow B$ is a if and only if (iff). I'm not sure how it fits in the discussion.

 

[Gavran]

The above $A \Leftrightarrow B$ is a if and only if (iff). I'm not sure how it fits in the discussion.

The Axiom of Choice and the existence of a Hamel basis imply each other.

 

[cianfa72]

The Axiom of Choice and the existence of a Hamel basis imply each other.

Not sure to get the point. From the iff implication the existence of Hamel basis is equivalent to AC is True.

Maybe the key point is that claiming AC is True or False is a different matter from excluding it from the underlying set theory employed. In the latter case, let me say, we can say nothing about it (neither True or False since it doesn't exist at all).

Does the above make sense ?

 

[nuuskur]

The argument is if AC is false, then the statement "every vector space contains a minimal generating subset" is also false. Therefore, there exists (at least one) a vector space such that none of its linearly independent subsets are generating. .. but can we find an explicit example of one?

 

[Gavran]

The answer to the post #8: Yes, it does.

 

[cianfa72]

But $\mathbb R_\mathbb Q$ is an infamous example of one where you can't get a basis without choice, but that doesn't yet imply it has no basis.

Ah ok. Basically your are saying that to pick a basis for $\mathbb R_\mathbb Q$ without assuming AC as True you can't. However that doesn't imply a basis for it doesn't exist at all.

How can we be sure that a basis for $\mathbb R_\mathbb Q$ actually exists?

 

[nuuskur]

Ah ok. Basically your are saying that to pick a basis for $\mathbb R_\mathbb Q$ without assuming AC as True you can't. However that doesn't imply a basis for it doesn't exist at all.

How can we be sure that a basis for $\mathbb R_\mathbb Q$ actually exists?

If AC is false, then there's no guarantee this is one of those vector spaces without a basis. Of course it's clear, a generating subset has to be uncountable, because countable subsets can only ever span countable subsets over $\mathbb Q$. But now you're in the uncountable territory, it's very difficult to do much of anything without (some form of) choice.

 

[cianfa72]

If AC is false, then there's no guarantee this is one of those vector spaces without a basis. Of course it's clear, a generating subset has to be uncountable, because countable subsets can only ever span countable subsets over $\mathbb Q$. But now you're in the uncountable territory, it's very difficult to do much of anything without (some form of) choice.

Sorry, the point I'm having trouble in understanding is that, I believe, $\mathbb R_\mathbb Q$ for instance either has or has not an Hamel basis, regardless of whether the Axiom of Choice is included in underlying set theory or not.

A different matter is to give a proof of the fact that $\mathbb R_\mathbb Q$ has an Hamel basis.

 

[nuuskur]

Yes, a statement is either true or false. The best we have in this instance is that it is consistent with ZF that R over Q does not have a basis.

In similar vein, it is consistent with ZF that every subset of R is Lebesgue measurable (as a consequence of R being a countable union of countable sets). Is it actually true? Who knows, but if it was true, then you won't get into trouble. It is not possible to exhibit a non measurable subset without (some kind of) choice.

 

[cianfa72]

Yes, a statement is either true or false. The best we have in this instance is that it is consistent with ZF that R over Q does not have a basis.

Let me check my understanding. Consider the statement "R over Q doesn't have an Hamel basis". Is it true ? Who knows, what we can say is that with ZF (not including AC) one won't be able to exhibit an Hamel basis for it. In this sense "R over Q doesn’t have an Hamel basis" is consistent with ZF.

In similar vein, it is consistent with ZF that every subset of R is Lebesgue measurable (as a consequence of R being a countable union of countable sets). Is it actually true? Who knows, but if it was true, then you won't get into trouble. It is not possible to exhibit a non measurable subset without (some kind of) choice.

Likewise, since without AC one isn't able to exhibit a non measurable set, the sentence "every subset of R is Lebesgue measurable" is consistent with ZF (AC nor included).

 

[cianfa72]

Sorry, does what discussed above have anything to do with Godel's incompleteness theorems?

 

[nuuskur]

I don't think it does, Gödel's theorem is about consistency and completeness, roughly stating that you can't have both if your theory is "rich" enough (ZF is rich enough). Regardless, if there was some way to demonstrably show in ZF alone that R over Q (or some other weird vector space) has no basis, then it would also have to be true in ZFC, which is a contradiction.

 

[cianfa72]

I don't think it does, Gödel's theorem is about consistency and completeness, roughly stating that you can't have both if your theory is "rich" enough (ZF is rich enough). Regardless, if there was some way to demonstrably show in ZF alone that R over Q (or some other weird vector space) has no basis, then it would also have to be true in ZFC, which is a contradiction.

Just to reiterate what you said. Suppose in the context of ZF set theory (i.e. without throwing in the AC) there was some way to prove that R over Q has no Hamel basis.

Then, since by including the AC axiom (taken as True of course) turning ZF into ZFC can be proven the existence of Hamel basis for R over Q, there would be a contradiction (since the statement "R over Q has an Hamel basis" can be either True or False regardless of ZF or ZFC.)