Algebraic Structures Applicable to All Sets

This exploration is inspired by the fact that, given the Axiom of Choice, any set can be given a group structure. (Fine print: throughout the rest of this post, we assume all sets are nonempty.) In fact, these two statements are equivalent. Questions about similar statements have popped up in other contexts; for example, in Topology as an Algebraic Structure, the question arises as to whether any infinite set can be given a field structure. (Clearly, not every finite set can be given a field structure, since we have the classification of finite fields.)

Here, we investigate these and similar questions. The general form of such questions is: given a universal algebraic signature $\Sigma$ and a set of $\Sigma$ -axioms $A$ , can any set be given a collection of operations on it of type $\Sigma$ satisfying $A$ ? For which $(\Sigma,A)$ would this be equivalent to the Axiom of Choice? (As we can see from the infinite versus finite point for fields, this is not the exact general form, but rather a rough source of inspiration.)

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First, if we look at the proof of equivalence of AC to general group structure applicability in its Wikipedia article, it seems that general cancellative magma structure applicability is sufficient to yield the Axiom of Choice. Thus, we can say that the following are equivalent:

The Axiom of Choice
General group applicability
General cancellative magma applicability

This is actually quite interesting, since it implies that given that any set can be turned into a cancellative magma, any set can then be turned into a group, which has a much richer structure.

What is the “richest” structure for which general applicability would be equivalent to the Axiom of Choice? That would stretch this kind of implication further.

Clearly, it can’t be a field. Is there a similar restriction on the size of a finite ring? No, there isn’t, since $\mathbb{Z}_{n}$ for any $n$ is a ring. In fact, $\mathbb{Z}_{n}$ for any $n$ is a commutative ring. Hence, any finite set can be given the structure of a commutative ring.

What about infinite sets? Can any infinite set be given the structure of a commutative ring?

For the sake of possible use in the topology exploration mentioned earlier, I would hope to show that in fact any infinite set can be given a field structure. If that’s the case, then that would automatically imply that any infinite set can be given a commutative ring structure, which would then yield that the following are equivalent:

The Axiom of Choice
General cancellative magma applicability
General commutative ring applicability

This would be quite awesome if it were true!

Actually, if we look again at the Wikipedia article, this time at the proof that AC implies general group applicability, we see that, if we can impose an algebraic structure on the set $F$ of finite subsets of a set, then we get general applicability of that algebraic structure to infinite sets. The group structure comes from symmetric difference on $F$ . However, if we take symmetric difference and intersection, we can see that we can turn $F$ into a Boolean ring. Thus, given AC, we actually have general Boolean ring applicability for infinite sets. The converse is clear too, since every Boolean ring is a group (and every finite set is already a group.) Furthermore, since every Boolean ring is commutative (as can be seen from its Wikipedia article), we yield that the following are equivalent:

The Axiom of Choice
General cancellative magma applicability
General commutative ring applicability
General Boolean ring applicability for infinite sets

Inclusion of this last statement is a result not encompassed by our infinite-field conjecture (at least not directly), since fields aren’t generally Boolean rings. As mentioned, any of these statements is in turn equivalent to general applicability of any algebraic structure “between” cancellative magma and commutative ring, i.e., any class of algebras that is between the classes of cancellative magmas and commutative rings (ordered by class inclusion.)

From the Wikipedia article on Boolean rings, a finite Boolean ring must have cardinality a power of two, so general Boolean ring applicability isn’t true.

Let’s set about trying to prove our infinite-field conjecture now:

Conjecture. Given AC, any infinite set can be turned into a field.

Attempt. We know at least that for the smallest cardinality, $\aleph_{0}$ , we have the field $\mathbb{Q}$ . Inspired by the Generalized Continuum Hypothesis, is it possible to show that if there is a field of cardinality $c$ then there is a field of cardinality $2^{c}$ ?

This can be good for a first foray, but we’d have to see what else we’d need too to achieve our full result.

Let’s try this for now though; at least we would establish the result for any set of cardinality $\aleph_n$ for some $n$ . We want to show that: given a field $K$ , there is a field of cardinality $2^{|K|}$ , or a field in bijection with the power set of $K$ . Thus, we want to show that: given a field $K$ , there exists a field in bijection with $2^K$ .

Can we do something similar to how we went from $\mathbb{Q}$ to $\mathbb{R}$ ? Two possibilities I can think of are to use construction by “decimals” (look at my exploration on this) or something similar to the Exodus reals. However, we have to be careful, since we know that any Dedekind-complete ordered field must be isomorphic to $\mathbb{R}$ and thus have cardinality $\aleph_{1}$ , limiting applicability to other cardinalities.

It remains to figure this out. I am also currently learning about cardinal numbers, transfinite induction, etc., and I will see if those can apply to this problem.

This discussion is continued in some follow-up posts: one post discusses in more detail the general field applicability to infinite sets, while another post discusses a universal algebraic formulation of the general problem.

Modified February 21, 2023, to correct some oversights and mistakes.

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