The axiom of choice is, for ZF, equivalent to the statement that every vector space has a basis. The implication of AoC by the existence of a basis for any vector space is shown in this paper.

The proof of that theorem can be summed up as following: One adjoins all members of the sets $X_i$ from which one needs choice as variables to an arbitrary field $k$ creating the field of rational functions $k(X)$ over that field. Then a subfield $K$ of that field is defined through a variation of homogeneity that considers the index of the $X_i$ from which variables are occurring and finally the span of the variables as a vector space $V$ over $K$ is considered. A basis for $V$ over $K$ is chosen, and the construction is such that for all $x \in X_i$, the same set of basis elements occurs with non-zero coefficient in their $K$-representation. This fact is then used together with a similarity of the coefficients occuring to pick a unique $x_i \in X_i$ and an element was chosen as desired.

Now, this comment made me curious if there can be a $\mathbb{Q}$-basis of $\mathbb{R}$ when choice fails at size continuum or below.

That separates into three potentially hard (or, perhaps more likely, easily negatively answered) questions:

a) Can we conclude anything from the existence of one specific basis? The construction of the critical field in the proof relies on the structure of the set family we choose from, and so it seems that these fields will not be isomorphic for different families of sets. But that doesn't say there cannot be a different proof where they are. Or it might also be possible to choose that one guaranteed basis for a family such we can imply choice for all other set families of the same cardinality - but then, does such a family exist? Any direct proof of the existence of well-orderings I know use multiple instances of choice, but perhaps one carefully chosen one can be enough?

b) Assuming a) is positively answered, can we make any conclusions about lower cardinalities? Or, under ZF more strongly, can we actually say something about all subsets even if they are not comparable - can they still exist if we have choice for all families of sets of cardinality $\frak c$?

c) Even if a) is answered positively, could a $\mathbb{Q}$-basis for $\mathbb{R}$ be used as this basis? This would then obviously require a different proof.

To make the question precise, consider three versions of "hold up to $\frak c$":

Do we have choice for families of sets where both the index set and all sets are subsets of $\mathbb{R}$?

And, not necessarily equivalently, do we have choice for families of sets where both the index set and all sets have defined cardinalities $\le\frak c$ or perhaps $<\frak c$?


1 Answer 1


According to the Consequences of the axiom of choice website it is not known whether a Hamel basis for $\mathbb R$ over $\mathbb Q$ implies $AC(\mathbb R)$ (choice function on $\mathcal P(\mathbb R)\setminus{\varnothing}$). The latter is equivalent to the assumption that $\mathbb R$ can be well ordered.

To search those, 367 is the form number of a Hamel basis and 79 for the other one.

I find the question a little bit illformed. Why? We can violate the axiom of countable choice in ranks so high that the real numbers are unaffected by this, Monro showed that we can add Dedekind-finite sets which can be mapped onto $\kappa$ for an arbitrarily high $\kappa$, we do that by adding subsets to $\kappa$ instead of adding real numbers.

Suppose we start with $V=L$, in which $\frak c=\aleph_1$, and we add a Dedekind-finite subset in rank $\aleph_{\omega_1}^+$. This rank is so high that there is nothing about the real numbers which did not already occur at this stage. On the other hand, we no longer have countable choice!

To further the divide there are several different weakenings of the axiom of choice which are non-equivalent.

  1. There is $AC_\lambda$ which asserts that families (of nonempty sets) of size $\leq\lambda$ have choice functions;
  2. There is $W_\lambda$ which asserts that every cardinality is comparable with $\lambda$; and
  3. There is $DC_\lambda$ which asserts that for every binary relation on a nonempty $S$ if every "chain" of length $<\lambda$ has an upper bound then there is a chain of length $\lambda$.

Between the three there is only one real implication, given a regular $\lambda$, $DC_\lambda$ implies both others. They do not imply each other except some minor case of limit cardinals where we have $AC_{\operatorname{cf}(\lambda)}$ and $DC_{<\lambda}$.

This makes the question "do we have choice up to $\frak c$?" somewhat ill-formed, even more since we don't know if $\frak c$ has to be well-ordered to have a basis; if it is not well-ordered then $AC_\frak c$ has an even weirder formulation.

There is also all the issue of other forms of choice: Boolean Prime Ideal Theorem (and its many equivalents) and so on which might imply somehow that a basis exists. These are long known to be highly separated from any of the three above (Cohen's first model: countable choice fails in the real numbers, but BPIT holds).

  • $\begingroup$ While I definitely find all those facts interesting, I'm not sure I understand why the question is ill-formed. Would it be more precise if I stated this as two different questions such as "given the assumption, do we have a choice functions for any family of sets where the index set and each set are subsets of $\mathbb{R}$" or, alternatively (which is part of the question), with defined cardinality $\frak c$ or lower for both index set and the sets in the family? What I mean to ask is for both index and the sets in the family to be "below" $\frak c$, but I realize "below" is not clear. $\endgroup$
    – Desiato
    Mar 21, 2012 at 12:47
  • 1
    $\begingroup$ Yes, this would be much clearer. Because the value of $\frak c$ is independent of ZFC (as an $\aleph$ cardinal), it is not a well defined question whether your mean $AC(\mathbb R)$ holds or do you mean that $\mathbb R$ can be well-ordered and $AC_\frak c$ holds. As for "below" this is even worse, suppose there was a Dedekind-finite subset below $\mathbb R$ then there are infinity of them; and those can become wildly and uninhibited in the sense that we don't really understand them like we do $\aleph$ cardinals, and they obey infinitary definitions much worse than we are used to in ZFC. $\endgroup$
    – Asaf Karagila
    Mar 21, 2012 at 13:04
  • $\begingroup$ Yes, as noted, I realized "below" is not well defined. The thought I originally had was actually that the existance of such a basis might, perhaps, prevent the existance of such "pathologic" sets "below", but given your reply that seems increasingly unlikely. $\endgroup$
    – Desiato
    Mar 21, 2012 at 13:27
  • $\begingroup$ I'm going to accept this answer, since the very first statement essentially says this is an open problem. I wonder if it'd be appropriate to add the open-question tag? $\endgroup$
    – Desiato
    Mar 24, 2012 at 17:49
  • $\begingroup$ @Desiato: I gave it some thought but couldn't solve it. I will be quite surprised if the existence of a Hamel basis would imply a well ordering of the real numbers. $\endgroup$
    – Asaf Karagila
    Mar 24, 2012 at 19:38

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