General Field Applicability to Infinite Sets

In this post, we undertake a more focused investigation into the conjecture that every infinite set can be turned into a field, as stated in this post. (We assume the Axiom of Choice throughout.)

For a cardinal number $c$ , let $P(c)$ be the assertion that a set of cardinality $c$ can be turned into a field.

Known facts that we gather from online:

Cardinals are well-ordered.
While many of the statements I have found online defining transfinite induction state that it applies to any well-ordered set, based on some of its applications (e.g. to ordinals) it seems to be the case that transfinite induction applies to any well-ordered class. We will consider this a fact here. (Ordinals don’t form a set, but rather a proper class.)
- Hence, transfinite induction applies to cardinals.
  - (The distinction between class and set is necessary for this conclusion, since the cardinal numbers don’t form a set, but rather a proper class.)
We have $P\left( \aleph_{0} \right)$ and $P\left( \aleph_{1} \right)$ .

Ideas:

Assume that we show that: if , then for any with . Does this yield our desired result? Does this at least yield for all non-strong limit cardinals ?
- We attempt to proceed by transfinite induction. Let be a cardinal, and assume that for any with we have .
  - Say that we can produce a cardinal whose successor is . In that case, we have , thus is true. However, by definition of , since is its successor, , thus we have .
    - Can we always produce such a cardinal? According to successor cardinal (planetmath.org), this is only possible if $d$ is a successor cardinal. So this doesn’t reach strong limit cardinals.
    - Can we modify this to reach all cardinals? We’d like to produce a cardinal $c^{'}$ with $c^{'} < d \leq 2^{c^{'}}$ , then we’d be done. But from the definition of strong limit cardinal, if $d$ is a strong limit cardinal then if $c^{'} < d$ we’d have $2^{c^{'}} < d$ . Thus, this strategy can’t work for strong limit cardinals.
  - Say we have a set of cardinality . The induction assumption implies that any infinite subset of of smaller cardinality can be endowed with a field structure. Can we “put these structures together” to yield a field structure for ?
    - We can put $S$ in bijection with “ $d$ copies of $\mathbb{Q}$ .” In other words, we can put $S$ in bijection with $S^{'} = \cup_{s \in S}\left\{ (s,q):q\mathbb{\in Q} \right\}$ . (This is because the set would have cardinality $d\aleph_{0}$ , which is equal to $d$ .) Can we show that an arbitrary union of fields (or even just an arbitrary union of sets bijective to natural numbers) can be given a field structure?
    - As preliminaries, we can endow $S$ with a Boolean ring structure (as has already been seen.) Thus, we have $+ , \times ,0,1$ on $S$ .
    - For , how can we achieve a field structure?
      - Let’s try: given $(s,q),(t,r) \in S^{'}$ , we can define $(s,q) + (t,r) = (s + t,q + r)$ , $(s,q) \times (t,r) = (st,qr)$ . Does this work?
        
        An equivalent way of asking this is: is the Cartesian product of a Boolean ring and a field again a field? We clearly have $(0,0)$ as an additive identity and $(1,1)$ as a multiplicative identity. Is it true that any element other than $(0,0)$ is invertible? Consider $(s,0)$ ; is this invertible? Does there exist $(t,r)$ with $(s,0) \times (t,r) = (1,1)$ ? By our definition, this would imply $0r = 1$ , which is impossible for $r$ rational. Thus, this doesn’t work.

Started February 21, 2023.

—

EDIT (03/28/2023): A resolution to this question in a vastly general way is given by the Löwenheim-Skolem theorem, referred to in this post on a general universal algebraic formulation.

Share and like (need WordPress account to like):

Leave a comment Cancel reply