# What's an example of a vector space that doesn't have a basis if we don't accept Choice?

I've read that the fact that all vector spaces have a basis is dependent on the axiom of choice, I'd like to see an example of a vector space that doesn't have a basis if we don't accept AoC.

I'm also interested in knowing why this happens.

Thanks!

• Just a guess: the real numbers as a vector space over the rational numbers?
– bof
Dec 23, 2015 at 19:00
• Dec 23, 2015 at 23:05

Classically, they can be pretty simple: that is,

We can have a model $M$ of ZFC, with an inner model $N$ of ZF, such that there is a $\mathbb{Z}/2\mathbb{Z}$-vector space $V\in N$ such that $(i)$ $N\models$"$V$ has no basis" and $(ii)$ $M\models$"$V\cong\bigoplus_{\omega}\mathbb{Z}/2\mathbb{Z}$".

Of course, inside $N$ this characterization of $V$ won't be visible.

I almost forgot the classic: $\mathbb{R}$, as a vector space over $\mathbb{Q}$! I'd argue this is "more complicated" than the one above in certain senses, but in others its more natural.

As to why this happens: basically, consider a "sufficiently large" vector space $V$ with lots of automorphisms. Then, starting in a universe $M$ of ZFC which contains $V$, we can build a forcing extension $M[W]$, where $W$ is a "generic copy" of $V$. That is, $W$ is isomorphic to $V$, but all twisted around in a weird way. Now, we can take a symmetric submodel $N$ of $M[W]$ - this is a structure between $M$ and $M[W]$, consisting (very roughly) of those things which can be defined from $W$ via a definition which is invariant under "lots" of automorphisms of $W$ - specifically, invariant under every automorphism fixing some finite set of vectors! But as long as $W$ is sufficiently nontrivial, no basis (or, in fact, infinite linearly independent set) is so fixed.

Of course, I've swept a lot under the rug - what's a forcing extension? what exactly is $M[G]$? and why does it satisfy ZF? - but this is a rough intuitive outline.

Actually, in a precise sense, this is the wrong answer: I've just argued that it's consistent with ZF that some vector spaces not have bases. But, in fact, Blass showed that "every vector space has a basis" is equivalent to the axiom of choice! See http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf, which is self-contained. Blass' construction actually proves that "every vector space has a basis" implies the axiom of multiple choice - that from any family of nonempty sets, we may find a corresponding family of finite subsets (so, not quite a choice function); over ZF this is equivalent to AC (this uses the axiom of foundation, though).

Blass argues roughly as follows. Start with a family $X_i$ of nonempty sets; wlog, disjoint. Now look at the field $k(X)$ of rational functions over a field $k$ in the variables from $\bigcup X_i$; there is a particular subfield $K$ of $k(X)$ which Blass defines, and views $k(X)$ as a vector space over $K$. Blass then shows that a basis for $k(X)$ over $K$ yields a multiple choice function for the family $\{X_i\}$.

So now the question, "How can some vector spaces fail to have bases?" is reduced (really ahistorically) to, "How can choice fail?" And for that, we use forcing and symmetric submodels (or HOD-models, which turn out to be equivalent but look very different at first) as above.

• This is a nice answer, but I don't see how the part after last horizontal line is relevant. Dec 23, 2015 at 19:18
• @Wojowu I'm trying to answer the "why". There's two different questions there: why it can happen, and why it must happen under certain circumstances (specifically, if choice fails). Your answer shows why it can happen (assuming we believe the consistency of "all sets are measurable"), but says nothing about the case where the first failure of choice is of very high rank (you mention this in your first sentence). The last bit of my answer shows (modulo "choice = multiple choice") why it must happen, if AC ever fails. Dec 23, 2015 at 19:23
• @Wojowu Also, Blass' argument is constructive: given a failure of multiple choice, he builds an explicit vector space without a basis. So I think it's also a good answer to the first part of the question. (My use of "constructive" here is soft - I'm not sure e.g. that Blass' analysis of the vector space's properties does not rely on LEM. I just mean that his proof builds examples nicely.) Dec 23, 2015 at 19:26
• I know this might be a bit late, but the symmetric forcing argument you outlined seems very interesting and I am currently trying to reconstruct it. Following similar steps it should be possible to construct for any field $K$ a model of ZF in which the space $K^\mathbb{N}$ and any space containg it have no basis over $K$. However I couldn't find any source for this type of construction. So do you happen to know a source for the argument presented in the answer above where a symmetric forcing extension is created by attaching a generic copy of some vector space? Dec 4, 2020 at 18:27

If we only assume $\sf\neg AC$, then there is no hope for us to find a specific example, because it might be that such vector spaces have cardinalities well beyond what we can describe.

However, if we assume some stronger negation of $\sf AC$, for example axiom of determinacy, then one example would be $\Bbb R$ considered as a vector space over $\Bbb Q$.

Of course, $\sf AD$ is a bit of an overkill here. Nonexistence of Hamel basis for $\Bbb R$ follows already from "all sets of reals are measurable", and I believe even "all sets of reals have Baire property", the latter being equiconsistent with $\sf ZF$. Hence it is consistent with $\sf ZF$ that $\Bbb R$ as a vector space over $\Bbb Q$ has no basis.

As for "why this happens", let me show that under measurability assumption there is no such basis. Under this assumption, it's clear that every function $\Bbb R\rightarrow\Bbb R$ is measurable. Now it can be proven that every measurable function $f:\Bbb R\rightarrow\Bbb R$ which satisfies $f(x+y)=f(x)+f(y)$ for every $x,y$ must be in fact linear. This is a result due to Sierpiński I believe (I would be grateful if someone has posted a reference in a comment). However, if we had a Hamel basis, we could easily construct a function satisfying this condition but not linear: we can arbitrarily assign values of $f$ to elements of the basis and then uniquely extend it to whole $\Bbb R$, so e.g. if we define $f$ to be zero on all but one element of the basis, we get a desired function, which we, however, have proven can't exist.