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I know that we need the axiom of choice to deal with infinite collections of non-empty sets, but do we care if the infinite is countable or uncountable?

Here is the quote from Arturo Magidin "let's say that for a family of nonempty sets indexed by a natural number you do not need the Axiom of Choice to get a choice function, and this can be shown by induction on the index set"

Picking from an Uncountable Set: Axiom of Choice?

My question is:

1. Isn't the set of natural numbers is countable infinite? How is this infinite does not require the axiom of choice? Or the infinity that we talk about is always uncountable infinite?

2.The induction can go through all the natural numbers, we don't think of this as infinite?

3. Does such notation $\infty$ $$\sum^{\infty}_0$$ always means uncountable infinite? Then, does any distinct notation represent countable infinite?

4 Is this sum mean countably infinite sum $$\sum_{i\in\mathbb{N}}$$

LJNG
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  • https://en.wikipedia.org/wiki/Axiom_of_countable_choice – Vivaan Daga Sep 26 '21 at 12:01
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    Induction is used to prove that something is true for each natural number. Not that it is true for the set of natural numbers. – Asaf Karagila Sep 26 '21 at 12:15
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    @PaxDaga: Picking from a set does not require choice; whether the set is finite or infinite is irrelevant. And picking form sets finitely many times does not require choice. The problems arise when infinitely many "pickings" are required, and it has nothing to do with the size of the sets in question. – Arturo Magidin Sep 27 '21 at 02:37
  • @ArturoMagidin Of course for any set one can use existential intitiation that's what I meant but it was phrased incorrectly. I meant that exactly what you said in that post... – Vivaan Daga Sep 27 '21 at 02:47
  • @ArturoMagidin Any how I meant finite family but in order to avoid further confusion ill deltelte the comment – Vivaan Daga Sep 27 '21 at 02:54

1 Answers1

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First of all, Arturo Magidin refers to a family of nonempty sets indexed by a natural number, and not the natural numbers. What is meant here, is a family indexed by the finitely many elements $\{0,1,\dots,n-1\}$. This is the standard way to define natural numbers in set theory.

The Axiom of Choice states that every family of nonempty sets has a choice function, but this does not imply that there do not exist (infinite) families of nonempty sets that have a choice function without needing the Axiom of Choice. If I take the sets $X_n=\{2n,2n+1\}$ for each natural number, then the function $f:X_n\mapsto 2n$ is a choice function on $\{X_n\mid n\in\Bbb N\}$, even though this is an infinite family.

However, it is consistent with the failure of Axiom of Choice that there are countable families of nonempty sets that do not admit a choice function, so induction is not going to help: natural induction works, which is to say, we can prove that every family of size $n+1$ has a choice function from the assumption that every family of size $n$ has a choice function; but we cannot get transfinite induction to work, since the limit step fails.

It is similar to how we need an Axiom of Infinity to prove the existence of an infinite set, even though we can prove the existence of arbitrarily large finite sets without it. So to say, induction is not strong enough to create infinite objects out of finite ones.

This generalises, in fact, to higher cardinalities. If we have a choice function for every countably infinite family of nonempty sets, this does not imply that there exists a choice function for every uncountably infinite family of nonempty sets.

As for 3. and 4., naturally it depends on context what these notations mean.

Vsotvep
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    https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers is worth mentioning I think – Vivaan Daga Sep 26 '21 at 12:15
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    @PaxDaga added :) – Vsotvep Sep 26 '21 at 12:16
  • Thank you for your reply. For a collection of not well-behaved(defined/ordered) nonempty sets, either collection is countable or uncountable infinite, we need the AC? – LJNG Sep 26 '21 at 13:39
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    @LJNG It is not just as simple as a set that isn't ordered (I'm not sure what a "non-defined set" would be); families of set that don't admit a choice function can be quite subtle, and reversely families that do admit a choice function can be quite 'non-standard' compared to our conception of sets with choice. – Vsotvep Sep 26 '21 at 23:33
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    For example, each set in a family may have cardinality $2$, and thus for each set there exists a (well-)order of length 2, yet there may not exist a choice function on the collection of all sets in such a family. For example, consider the set $\mathcal P(\Bbb R)\times\mathcal P(\Bbb R)$, where each element is a pair of subsets of reals. How would you describe a choice function on this set? (in fact, it is consistent that such a choice function does not exist, which is the essence of the famous proof by Paul Cohen that the Axiom of Choice can consistently fail with respect to $\mathsf{ZF}$). – Vsotvep Sep 26 '21 at 23:38
  • @Vsotep It seems that your first example is indexed by natural numbers and that infinite collections do not require AC, a similar construction indexed by real numbers also does not require AC, so the claim "dealing with infinity requires AC" is not always true? – LJNG Sep 26 '21 at 23:49
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    @LJNG The indexing of the infinite set has not much to do with it. There's families of sets with any infinite indexing that fail to have a choice function, and ones that admit a choice function. The example of the $X_n$ from above admits a choice function even in the absence of the Axiom of Choice despite being infinite, and it is consistent that there is a countable family of pairs of subsets of the reals without a choice function, despite being countable. – Vsotvep Sep 26 '21 at 23:55
  • However, if there is a well-order over the whole domain of the family (i.e. the union of all sets in the family), then there is a guaranteed choice function (just take the minimal element with respect to the well-order). The Axiom of Choice is equivalent to the statement that such a well-order exists for any set. Weaker forms exist as well, such as any countable family admitting a choice function (and per the last paragraph of my answer, this does not imply the full Axiom of Choice) – Vsotvep Sep 27 '21 at 00:00