Here is my new visualization of the Axiom of Choice

[discountbrains]

I've never quite accepted the Axiom of Choice and have heard some others don't either. This is how I now think of it: For example, suppose you were going to play in a doubles tennis match and you were given a group of 50 people to choose your one partner from. You may have no idea of how good any of them are so you choose one at random. You are given the right or privilege by the Axiom of choice , but it doesn't tell you which one to choose. It's possible you may know which ones are the best yourself, but the Axiom of choice doesn't pick one for you. This could be repeated multiple times; you could be presented with many different sizes of groups or sets. The Axiom of choice only gives you permission to choose one.

[studiot]

The 'Axiom of Choice' refers to set (was the tennis reference a deliberate pun?) theory.

So you must first define your set.

The condition for membership of that set is that you must be a player in the match.

Nothing more.

So where is the a failure of the set axioms?

[mathematic]

The axiom of choice refers to infinite sets.  For finite sets, there is no problem in choosing an item.  For infinite sets, the question is forming a set made up of one item from each of the infinite sets.

[taeto]

On 6/29/2018 at 11:43 PM, mathematic said:

The axiom of choice refers to infinite sets.  For finite sets, there is no problem in choosing an item.  For infinite sets, the question is forming a set made up of one item from each of the infinite sets.

Your are right of course. However,  the classical finite example considers the, presumably finite, collection of pairs of socks in your closet. Assuming that any two pairs are different, say, they have different colors, but the two socks from each pair look identical. You ask a friend, for whatever reason, to go and pick one sock from each pair. You cannot really explain to your friend which one to pick out of each pair; to say "take the left one" doesn't work as they are identical. If your friend is compliant and the closet not a jumbled up mess, they will be able to pick just one sock from each pair, and in the end you have a well-defined collection of different looking socks. The axiom of socks, sorry, choice, acts similarly, by allowing you to pick a set similarly, without any need to explain how it is done by any rule of picking.

[mathematic]

The problem with infinite sets sets is that to get the choice set, you need an infinite amount of operations.  For non-countable sets, it seems impossible, so we have an axiom.

[taeto]

12 hours ago, mathematic said:

The problem with infinite sets sets is that to get the choice set, you need an infinite amount of operations.  For non-countable sets, it seems impossible, so we have an axiom.

That does not sound right. You can never perform an infinite amount of operations anyway. But it is quite possible to get a choice set from an uncountable set of uncountable sets, by just a single "operation". Example: the Euclidean plane \( R^2 \) is the union of disjoint uncountable sets \( R(x) := \{ (x,y) : y \in R \} \). The graph of \( y = x \) is the uncountable choice set \( \{ (x,x) : x \in R \} \) of the sets \( R(x), \)which you get by application of a single axiom in ZF, no choice axiom needed. It is only sometimes that you can only get the choice set by using AC. 

[mathematic]

Your example is a particular case.  It does not address the general problem, where the number of sets is uncountable.

[taeto]

10 hours ago, mathematic said:

Your example is a particular case.  It does not address the general problem, where the number of sets is uncountable.

I do not understand your comments. Yes, I am aware that a single example is only one particular case. However the example shows that even if given an uncountable set of sets each of which is also uncountable, then it may be quite easy to point to a choice set. You appeared to suggest that this seems impossible, which appears to be a comment about the general case. I tried to point out that those cases when you must use AC are special cases, and not the general rule. I probably misunderstood your original comment. What I think you could have said is that to a naive person it might seem that you can always find a choice set, whereas to a sophisticated person it is clear that no choice set has to exist unless something strong enough like AC can be applied.   

Uncountability does not enter into it that strongly. As another example, assume that a relation ~ is defined on the set of real numbers \(R\) so that \( x \sim y \) holds whenever the difference \( y - x \) is a rational number, i.e. \( y-x \in Q. \) It is easy to check that  ~ is an equivalence relation, which means that \( R \) is partitioned into equivalence classes, each of the form \( [r] := Q + r, \) for some \( r \) that is irrational or zero. Every equivalence class is just a shifted copy of \( Q, \) so in particular it is a countable set of real numbers. Even though each set \( [r] \) is countable, it is impossible to prove the existence of a choice set for the equivalence classes without using AC. (A particular such choice set is called a Vitali Set and is important in measure theory.) 

Edited July 3, 2018 by taeto

[studiot]

Thank you for this interesting discussion gentlemen. +1 each.

[studiot]

On 29/06/2018 at 10:43 PM, mathematic said:

The axiom of choice refers to infinite sets.  For finite sets, there is no problem in choosing an item.  For infinite sets, the question is forming a set made up of one item from each of the infinite sets.

I initially read this to mean that the Axiom of Choice only refers to infinite sets.

I hope you didn't mean that as none of my texts say that and some give specific finite examples.

 

[wtf]

40 minutes ago, studiot said:

I initially read this to mean that the Axiom of Choice only refers to infinite sets.

I hope you didn't mean that as none of my texts say that and some give specific finite examples.

 

 

AC applies to finite collections of sets, but it's not needed for finite collections of sets. That is: If we have one nonempty set, we may choose an element from it, since the set's nonempty. If we have two nonempty sets, we may choose an element from each. We can always choose a an element from each of a finite collection of sets. (The individual sets themselves may be finite or infinite, makes no difference).

It's only in the case of an infinite collection of sets that we need AC to prove that we can simultaneously choose an element of each set. 

In other words AC applies to finite collections, but it's not needed. It's using dynamite  to kill a fly. Effective, but overkill.

Edited July 6, 2018 by wtf