Apologies for asking such a basic question, but I am trying to understand how Set Theory represents identical sets (in Quantum Physics sets of identical particles can behave oddly) and the question arose when I read about Russell's countably infinite set of pairs of Socks in various Axiom of Choice questions.

Instead of socks turn it into something mathematical, so represent a sock as the set for the number 1. So a pair of socks is represented as the set {1,1} and call this set P. This means that the countably infinite set of pairs of socks becomes {P,P,....}.

The axiom of extensionality is in Set Theory $\forall a \forall b (a=b \leftrightarrow \forall z (z\in a \leftrightarrow z \in b))$

This presumably means {1,1} = {1} and also {P,P,...}={P}.

Also presumably the cardinal number of any set of identical sets is always 1.

So it appears that Set Theory can't actually distinguish a pair of socks from a single sock when considered as sets, as Set Theory would always think each pair of socks is actually only one sock. So how can the Russell question about selecting one sock from each of an infinite pair of socks (either as a sequence of pairs or as a set of pairs) be formally represented in Set Theory in a form that is relevant to discussions about the axiom of choice ?


Russell's example is divulgative, in order to clarify the need of the Axiom of Choice when dealing with infinite sets :

Russell coined an analogy: for any (even infinite) collection of pairs of shoes, one can pick out the left shoe from each pair to obtain an appropriate selection; this makes it possible to directly define a choice function. For an infinite collection of pairs of socks (assumed to have no distinguishing features), there is no obvious way to make a function that selects one sock from each pair, without invoking the axiom of choice.

Socks are individual objects and not sets. Thus, the Axiom of Extensionality will not apply to them.

If they were sets, having no elements they will be all the same set, and specifically the empty set.

Thus, a pair of socks is a set with two "distinct" elements.

The issue is : from a mathematical point of view there is no "uniform" property that can identify in one fell swoop one of the two socks of each pair from the other one, in the case of an infinite collection.

This is not the case with the infinite collection of pairs of shoes, where we can simply say : "choose the right shoe".

  • $\begingroup$ After looking up what divulgative means I then followed up your point to see if Russell just used it as a way of trying to get the general idea across to the public about the limitations of distinguishing each element in an infinite set using a single finite logical statement. I have just obtained Russell’s book containing the sock example and sure enough after the sock example is presented it says “… and the case of socks, with a little good will on the part of the reader, may serve to show how a selection may be impossible”. So Russell's point is deep but the example is divulgative. $\endgroup$ – Little Cheese Sep 9 '18 at 8:56

The real problem here is that the sock analogy is a bad analogy that misses important features of how set theory works.

Yes, whenever you have two sets (in ordinary well-founded set theory), the axiom of extensionality tells you that they are distinguishable -- there will be something that is an element of one of them and not an element of the other one.

A better analogy would be, instead of speaking of infinitely many pairs of indistinguishable socks, you have infinitely many pairs of various woolen objects, and you want to instruct a stupid assistant to pick one object from each pair.

Whenever you look at one of the pairs, it is easy to see a difference between them. Some are red, some are blue. Some are big, others are small. Some are knit, others are crotcheted. Many are just random snarls of yarn.

However, whenever you ask your assistant to pick one from each pair, he will go working for some time, and then come back: "Boss, which of these two ones do you want? You haven't told me which one to pick." Each time you can find something to do with that pair, but the questions keep coming. Take the heaviest? Somewhere there's a pair of woolen things that weigh exactly the same. Take the one with the fewest stiches in total? After you've defined what counts as a stitch for a random snarl of yarn, you'll find that one of the pairs end up with the same count under that procedure. (Oh, but they are different colors! How did you forget to instruct him to pick the red one when that happens?)

The axiom of choice states that you have an assistant with sufficient initiative to just pick one of them already each time he comes to a pair that you haven't left instructions for distinguishing between.

  • $\begingroup$ Thanks for the Physicist friendly explanation. Looking at your kind explanation it looks like AC is further developing the properties of the Set Theory “Axiom Of Separation” – by extending it to some of the cases where a single (finite) logical statement can’t precisely describe each element in an infinite set. So it looks like AC is including a way of separating out infinite sets of elements that would otherwise be impossible using the Axiom of Separation ? $\endgroup$ – Little Cheese Sep 9 '18 at 9:16
  • $\begingroup$ @ColinHarling: Yes that's a way of looking at it. It may not be extremely informative, because the axiom of separation is already the swiss knife of set theory construction which we use for absolutely everything except the few specific cases it does not suffice for. It might arguably be more productive to think of it as an extension of the axiom of Replacement, avoiding the requirement that the formula we use to define replacements always points to a single value (but we still get exactly one replacement per element of the original set). $\endgroup$ – hmakholm left over Monica Sep 9 '18 at 20:13

So... you're living in a "one sock universe", where all the socks are the same? Or a "one ant universe", where all the ants are really just the same ant, moving really fast?

In mathematics two things are equal if they are the same. Two socks are not one sock. Two ants, even if you cannot distinguish which one is which, are not the same ant. Usually.

So you would represent one sock as $1$, but another sock as $2$ and another sock as $3$, and so on and so forth. But even that's wrong. Because that makes Russell's analogy odd, why can't you choose from pairs of socks? If the socks are labelled $1,2,3,4,5$ and so on, just choose the one with the smaller label from each pair.

The point is that Russell's analogy is based on the idea that you cannot label all the socks simultaneously. You can label all the pairs at the same time, yes. And you can distinguish between two given socks in the same pairs, but there is no feature that you can use that works to distinguish the socks in all the pairs at the same time.

  • $\begingroup$ Looking at your explanation (or was it many explanations …?), if the Set Theory “Axiom of Separation” is only allowed to use a single finite logical statement “P(x)=True” to describe each element in a separated out infinite set and “P(x)=False” if x isn’t in the set then AC seems to be about lack of definability in the Axiom of Separation. There will always be some sets which are indistinguishable using any finite logical statement P(x), which presumably can imply some elements of a set may be different (using an oracle) but incorrectly P(x) says they are the same ... divulgatively ? $\endgroup$ – Little Cheese Sep 9 '18 at 9:33
  • $\begingroup$ Sort of yes on the first, but surprisingly no to the "some sets will be indistinguishable". Definability is very subtle, and people don't pay as much attention to it as they should when making these kind of sweeping declarations. I have to admit, though, I'm not entirely clear as to what is your actual question here. $\endgroup$ – Asaf Karagila Sep 9 '18 at 14:09
  • $\begingroup$ Ah, I thought I had 'got' what Russell was saying, so looks like more thought is needed and I certainly hadn't appreciated that definability is so subtle. I was asking whether Russell was saying that using P(x) to attempt to separate a specific element from each of an infinite number of sets with two elements would not be possible because in one set with two elements, P(x) would return True for each element, thereby failing to be able to separate out only one of the elements, when creating the infinite series. But you say that isn't true. Any clues why ? $\endgroup$ – Little Cheese Sep 9 '18 at 20:21
  • $\begingroup$ Okay. The first thing to realize is that all these choiceless counterexamples are confusing and counterintuitive, until you develop very strong intuition about choice. Which most mathematicians (including many set theorists) do not really have. So the alternative is to work with the basic definitions, slowly, and so these type of arguments are often confusing for everyone. The second thing is that in that specific instance (or rather the formalization thereof), it is indeed the case that there is no predicate $P(x,u)$ (with parameters) that exactly one of each pair would satisfy. $\endgroup$ – Asaf Karagila Sep 9 '18 at 20:29
  • $\begingroup$ So to get the intuition I will allocate many more hours to learning about AC - to be able to explore a range of "lack of distinguishability" scenarios. To follow up your point about "definability is very subtle" any suggested references or an example to understand your point ? $\endgroup$ – Little Cheese Sep 10 '18 at 8:22

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