Bypassing the axiom of choice when there are infinitely many sets but there are constraints on the sets

Question

In the accepted response in this thread, Asaf Karagila wrote that

Note that it is perfectly possible to choose from infinitely many sets without the axiom of choice under a severe constraint that they have some common characteristic. From infinitely many sets of natural numbers we can choose the minimal in each set; from infinitely many finite sets of real numbers we can choose the maximal element of each set; etc etc.

For example, in the original post in this thread, Framate took the set $Y$ (which was not yet known to be finite) and selected from the preimage of every element $y\in Y$ the element with the least index using the well-ordering principle.

I have two questions:

1. Regarding the "severe constraints" that Asaf Karagila wrote about, can we say anything general about what such constraints allow for bypassing the axiom of choice?
2. One such severe constraint is that all of the sets are totally ordered and each has a least element in that total order. If there was just one such set, then it makes sense that we can choose its least element. But what rule exactly allows us to make this choice when there are more sets, say infinitely many? It seems like doing this requires some power, but I don't know the source of that power.

Answer 1

Your first question is poorly defined IMO, so I'm not sure how to answer it, but perhaps the answer to your second question will shed some light on the first.

Suppose $\{S_i\}_{i \in I}$ is an indexed family of subsets of $\mathbb{N}$. We can construct the mapping $m: I \to \mathbb{N}$ giving for each $i$ the minimal element of $S_i$ by constructing the set $m := \{(i,n) \in I \times \mathbb{N} \mid n \text{ is the minimal element of } S_i\}$. This perfectly doable without choice ans $m$ is a function under the set theoretic definition, since for each $i$, there will be a unique $n$ s.t. $(i,n) \in m$.

Answer 2

If $A$ is a non-empty set of natural numbers or a non-empty finite set of real numbers, we can write down the predicate $$\tag1y\in A\land\forall z\in A\colon z\ge y $$ It is a provable fact about non-empty subsets $A$ of $\Bbb N$ as well as about non-empty finite subsets $A$ of $\Bbb R$ that tehre exists one and only one $y\in A$ for which $(1)$ holds. This defines a choice function for these situations that does not require the Axiom of Choice. In other situations, you may resort to a different predicate that $(1)$, but only if you can prove the function property (i.e., uniqueness and existence of a suitable $y$).

Answer 3

The point of choosing from infinitely many sets simultaneously, without appealing to the axiom of choice, is that there is some common property to these sets which allows us to define—in the language of set theory, which is pretty meek—a unique element everywhere.

For example, if the sets we are given are all given with an associated group/ring/field/vector space structure, then all of these have an identity element $0$, which we can choose. Or, if all the sets are finite sets of real numbers, then we can choose the largest element from each set.

The simple answer is that there is a choice function if and only if we can find a function which assigns an element for each set. This is kind of vacuous and unhelpful, so instead we talk about structure and characteristics. Like being a group.

Caveat: Note that it is not enough that each set "can be endowed with a group structure", but rather we need to be given an assignment that gives all our sets this structure. Simultaneously. In other words, we need to have a choice function that chooses for each set the structure we want to use.

This is the source of this power. The source is that we sometimes know more about the sets, than just the fact that they are not empty. And since we are smart, we can use this knowledge to find choice functions. But not always.