Does whether or not we assume the Axiom of Choice change the definition of a set?

Question

A set is

... is a well-defined collection of distinct objects, considered as an object in its own right. For example, the numbers 2, 4, and 6 are distinct objects when considered separately, but when they are considered collectively they form a single set of size three, written {2,4,6} (from Wikipedia).

This means that $\{1, 1, 1, 1, 1, 1, ..., 2, 2, 2, 2, 2, 2, ..., 3, 3, 3, 3, 3, 3, ..., ...\}$ is the set $\{1, 2, 3, ...\}$, by definition. But, without the axiom of choice, how can we differentiate between the two?

Answer 1

The axiom of choice has nothing to do with the no repetition principle. In mathematics we do not define sets in terms of anything else but rather, when doing things rigorously, we describe the axioms that we want our sets to satisfy. There are different axiomatisations, and things get very delicate very quickly. One axiom, called the axiom of extensionality, expressed the desire for sets to be completely determined just by their elements. This entails that the sets you mentioned are equal. So, it is the axiom of extensionality that is at play here, not the axiom of choice.

The axiom of choice can be stated in several different ways and properly understanding its need and its consequences takes time. One way to think about it is by pretending you are going to be extremely precise about your sets and wonder, if you have lots of non-empty sets, what does it really mean to take one element from each set and form a new set from these elements. Well, if, for instance, all of those sets are sets of natural numbers, then you can simply choose the smallest element from each set. That's it - in a very short sentence I described precisely what that set is. Now, what happens if you don't have any mechanism for choosing an elements from each set? Well, suppose you have a million such sets. Well, you can write a definition consisting of a million lines, namely: choose an arbitrary element from the first set. Choose an arbitrary element from the second set. Choose an arbitrary element from the third set, etc., until you've chosen a million elements which will then form your new set. That is perfectly valid and does not require the axiom of choice. Note though, that it is a very long definition, but still finite. We do demand that our definitions, and our proofs, are all finite. So, what happens if you have infinitely many non-empty sets, and you wish to choose one element from each set and form those into a set? Well, now you have a problem since your 'definition' of this set will have to be infinitely long - and that is not allowed. This is where the axiom of choice comes in. It says that you can shorten this too long of a definition into a single invocation of the axiom of choice. Much like the principle of induction allows you to accept a recipe for infinitely many proofs into a single proof by invoking the principle of induction.

Answer 2

Yes, but not in the way you think it does.

The axiom of choice is not used to prove that $\{1,1,1,\ldots\}=\{1\}$. Only extensionality is used there. So your set is still just $\{1,2,3,\ldots\}$.

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The rest of this answer is going to be more technical. And you can ignore it, and just take the naive answer from above: no.

But now, let's talk turkey. Definitions are syntactic and sets are semantic. You can have sets which have one definition in one universe of set theory, and a whole other definition in another. And these definitions can use the axiom of choice. For example, $A=\{x\mid\mathsf{AC}\rightarrow x\neq x\lor\lnot\mathsf{AC}\rightarrow x=\varnothing\}$. In a universe where the axiom of choice fails, $A$ is $\{\varnothing\}$, but in a universe where the axiom of choice holds $A=\varnothing$. The set itself did not change, but the definition of we used to define this set gave us different results. So moving from a universe of $\sf AC$ to a universe of $\lnot\sf AC$ we had to pick a different definition in order to obtain the same set.

Answer 3

The axiom of choice doesn't come into play here. The axiom of extensionality says that "two sets are the same iff they have the same elements".

Your set $\{1,1,1,1,1,\ldots, 2,2,2,2,2,\ldots, 3,3,3,3,3,\ldots \}$ also has as its only elements $1$, $2$ and $3$, so it equals $\{1,2,3\}$; there aren't more "copies" of $1$ "to choose from" (if that is your choice connection..). In this notation, where we enumerate the elements, duplicates are not written, usually.

Answer 4

You're being confused by the difference between notation and what the notation stands for.

The ink blots (or pixels or whatever) that make up "$\{1,1,1,\ldots,2,2,2,\ldots, 3,3,3, \ldots\}$" are not what the set is -- they are merely part of a description of a set, and the set itself "doesn't know" how we have chosen to describe it.

What is a set, then?

Most fundamentally, a set is something we can ask, "is such-and-such one of your elements?" for every such-and-such we can think of, and get a "yes" or "no" answer back. Nothing more, nothing less.

The axiom of extensionality asserts that if we're looking at two such things and they agree about what their elements are -- that is if we ask each of them "is such-and-such one of your elements?" about the same such-and-such, then they give the same answers -- then they are really just the same set. We could perhaps imagine that the set has two different phone numbers we can call to ask what its elements are, but unknown to us both numbers lead to the same call center. But (so says the axiom) all other sets do know this, and they will give us the same answer for $\{555,1,2,3,4\}$ as for $\{555,4,3,2,1\}$.

The notation $\{1,2,3\}$ stands for

a set that answers "yes" to "is $x$ one of your elements?" if $x=1$ or $x=2$ or $x=3$, and "no" otherwise.

The notation $\{1,1,\ldots,2,2,\ldots,3,3,\ldots\}$ is informal but would stand for something like

a set that answers "yes" to "is $x$ one of your elements?" if $x=1$ or $x=1$ or $x=1$ or ... or $x=2$ or $x=2$ or $x=2$ ... or $x=3$ or $x=3$ or $x=3$ ..., and "no" otherwise.

For every possible $x$, these two descriptions both demand the same answer, so they describe the same set.