You're looking at it from a philosophical perspective where people care about whether the axiom is "true" or not. Other approaches to mathematics don't care, because they view its value not in some metaphysical notion of truth but in utility. They might say: this proof requires the axiom of choice and that proof doesn't, without assigning either one any value beyond utility (where utility might be, in some case, just their intellectual edification). I think this is what the article is trying to drive at as well. We can't "accept" or "reject" the axiom of choice based on some philosophical grounds, because those philosophical grounds would need to involve infinities [1], which, themselves, are not philosophical but mathematical. We either choose to use it or not based not on where we come from but on where we want to go.
[1]: The view that mathematical philosophy could be derived from core basic principles outside of mathematics was called Logicism, and was favoured by Bertrand Russell. Unfortunately, mathematical systems without some mathematical (i.e. non-"logical") axioms are quite limited, and almost all of them include at least one mathematical axiom, usually one that is equivalent to the existence of the natural numbers.
Isn’t this a philosophical perspective?
So it is a philosophical perspective…?
So one is not metaphysical because it is a tool to achieve ends?
(Sorry, if I come across as argumentative, I don’t mean to. I’m genuinely just curious.)
Or at least, that's the way I read it. The question of the value of metaphysics is an open one, which easily comes to dominate other questions if allowed to. I think the OP is saying, "I'm going to make a metaphysical commitment, and let other people worry about whether doing so is good metaphysics." (And there's good reason to think that it isn't -- but it's unclear whether that matters, or should matter.)
My apologies, but I think I’ve understood where I misread your comment. Thanks for being patient!
What's an example without any axioms? It doesn't seem like you can do anything without say ∀x: x=x (which I believe is reciprocal identity.)
It's Countable Choice, and of course the essence of Countability (why it's our favorite cardinality) is that almost all of our intuition about finite numbers can be correctly applied to countable infinities.
Uncountable Choice (a set for every Real number, a choice from every set) is where thing ago off the rails and ZF looks substantially different from ZFC.
a = the algorithm
A = set of all selection algorithms
b = any set
def a(x):
while True:
f = a(A) # you have to select an algorithm from the set of selection algorithms
result = f(x)
if result is None:
A = A - f # set of all selection algorithms without f
continue
else:
return result
a(b) #will always return a result or not terminate for any choice of b?
I'm looking at it and such an algorithm is defined in terms of itself (like a factorial, which is legal) and may never terminate (which also legal because an algorithm selecting the smallest number from all sets that only contain positive numbers will never terminate either).
Wait but then if you look at the code it provably will never terminate because A does not shrink as it recurses.
By _assuming_ that program terminates, you are yourself taking the axiom of choice.
If it _does_ terminate, then it's not an axiom anymore it's a proof.
The axiom of choice is an assumption that is neither known to be true or false.
Moreover, you'd probably want to limit the tries to algorithms that terminate. But that brings you into the halting problem.
I think us programmers think in terms of time. But in math there is no time so whether you do things in parallel or procedural is irrelevant. That's why you can discuss infinities in math.
Nope, you don't need the axiom of choice to define the sequence of all algorithms. The axiom of choice allows you to order any arbitrary set, but you don't need it for things you can construct an explicit order for, like the natural numbers. In the case of all algorithms, it's somewhat straightforward to construct the set of all of them. You can construct the sequence of all strings, right? You can construct them as "", "a", "b", "c", ... "y", "z", "aa", "ab", ... "ay", "az", "ba", "bb", ... Now, pick a programming language, like C. Given any valid string, you can determine if it is a valid C program in finite time, and if so you can convert it into a set of instructions to use in the dove-tailing procedure in finite time. Take the list of all strings in the manner described above. For each one interpret it as a C program, or if it's got invalid syntax interpret it as a program that immediately halts. Now you've got an enumerable sequence of algorithms. Since C is Turing complete you'll find every single algorithm in that sequence. There will be a ton of duplicates (for whatever notion of equivalence you want to use) but all the algorithms will be there, and in a well-defined order that you can enumerate through.
No. I'm saying selecting an algorithm out of the set of all algorithms.
I'm not saying defining the set of all algorithms.
Then you don't need to "randomly pick" an algorithm. You just start with the first algorithm in the sequence, and keep going.
That is still random. You are arbitrarily picking an encoding, (English) in this case. Why not a Russian programming language or Chinese? How did you *select* your encoding out of the set of all encodings?
The act of assigning order to an unordered set is arbitrary. ABC order is a made up concept. It's not numerical, it's an arbitrary language and an arbitrary order that's a by product of human culture. Thus invoking this is at it's essence invoking the axiom of choice. You are arbitrarily selecting an algorithm.
You can phrase things like Gödel did. For any encoding of algorithms into natural numbers which is bijective (ignoring invalid syntax), enumerate the algorithms as 0, 1, 2, 3, ... using the bijection, and then run them in parallel by taking steps [0, 0, 1, 0, 1, 2, 0, 1, 2, 3, ...] The user provides an encoding, without invoking the Axiom of Choice.
You only need the axiom of choice if you need to pick an element out of a set WITHOUT SPECIFYING which one you are picking. And sometimes not even then. For instance, if you want to prove all elements in a set have a certain property, often you will see proofs take the form "pick an element of that set, assume it doesn't have this property, then by X, Y, and Z we have a contradiction, thus all elements have that property". That doesn't involve the axiom of choice at all, even though a straight reading of that statement makes it sounds like we are making a choice. In truth that proof is saying we can do this with every single element of that set, so you aren't really making a "choice". You only need the axiom of choice when you are, say, stating the existence of a function without defining what that function is. For an example of an actual invocation of the axiom of choice, check out the proof sketch in the wikipedia article on Zorn's Lemma: https://en.wikipedia.org/wiki/Zorn%27s_lemma .
I can say arbitrary stuff like the set of all sets with positive numbers but when I say the set of all algorithms written in English and C++ suddenly I'm getting too specific. Where is the line drawn?
Algorithms themselves have no specific order. In order to define enumeration you must first start off by *selecting* which algorithm gets the first enumeration. This is a completely arbitrary choice.
And yes, which particular encoding you decide to use is arbitrary, but the point is that you can enumerate the set of all algorithms, and thus you can select one without needing the axiom of choice.
Try to select an algorithm out of the set of all algorithms without using an encoding. If you must use an encoding, please ensure that it's not a "particular" encoding.
You can't.
The point is all encodings in the known universe are "particular."
Additionally, to even use an encoding you have to *select* and encoding from the set of all encodings.
One way of selecting an algorithm is to select a way to encode the algorithms.
It's easy to see that this is true. You chose an arbitrary example above. Try to do the same without choosing anything arbitrary. You can't.
But I'd also add that you can have a choice function that isn't an "algorithm". An algorithm, at least in the sense I'd generally interpret the word, has finitely many instructions and at most countably many steps. If we have uncountably infinitely many uncountably infinite sets, it is possible to have a choice function that can't be described in a finite algorithm.
Like, think about a well-ordering of the reals. If you believe the axiom of choice, then one exists. But you can't tell me what it is, because that would involve handing me infinite amounts of information. And similarly you can't write down an algorithm to produce it, without writing down infinite amounts of data.
No my algorithm above will never terminate. There's a recursive call where the input never shrinks and it won't converge on a base case.
It's equivalent to saying
def f(x):
return f(x)
Basically ketralnis is clarifying context which is sort of lost with my post. There's an axiom and do you believe in the axiom of not?
If the axiom can be proven then it is not an axiom. But a theorem. What I'm doing here is kind of pointless because we know it's an axiom by definition, the question is whether this axiom is valid or not. Attempting to prove the axiom is a a signal that I'm lacking clarity with the logic here.
def a(x):
return a(x)1. Find any solid object.
2. Divide that object in infinite, dimensionless fragments.
3. Make as many copies as you want with the infinite fragments.
4. Reform the original object with the remaining fragments.
... I still have fragments please help.
Point (2) is fundamentally wrong in a critical way. The whole point of the Banach-Tarski Theorem is that you only have finitely many pieces. Of course these pieces are composed of infinitely many points, but it's a finite number of "pieces".
Others may find it amusing, sense of humour vary widely. Usually I just shrug and move on, but in the essence of this "joke" you've genuinely misrepresented the entire point.
This theorem is often quoted as "mathematically-proven way to deconstruct a sphere and build two identical spheres", but this can't translate to reality in any way, it's an abstract thought exercise. Of course, "mathematically-proven way to deconstruct a sphere and build two identical spheres in your imagination" doesn't sound as cool.
We model spheres as collections of points, and given a collection of points there is a fairly obvious way to model the idea of moving that collection around in a way that reflects moving a physical object. The problem is, given a sphere, we can partition the points into six sets, move those sets around, and recombine them to give us two spheres the same size as the initial one. In a very real sense, in this model we can "cut up" a sphere, then rearrange the pieces to make two spheres. We've doubled our volume.
Clearly this is nonsense, and it shows the limitations of the model. But equally, it tells us something important about the maths we use every day to model buildings, bridges, fluid flow, and more.
All models are wrong, some models are useful. -- George E. P. Box
I would add:
Some things are nonsense, but sometimes the nonsense can tell us useful things about the way the models are wrong.
I don't know the proof but there could be no algorithm like that for real numbers.
Why would a set of "negative numbers" not have a smallest element?
Negative numbers are also ordered in a number line (-5 is smaller than -2 for example).
So you can have a set like "all real numbers greater than 0". It doesn't have an obvious first element, even though you can always compare any two numbers and say "this one is less than that one".
There may be a non-obvious first element, i.e. some strategy for picking a first element that applies to all sets. That's the Well-Ordering Principle. Which feels false to most mathematicians, but it's equivalent to the Axiom of Choice, which feels true.
They are orderable ... For all x, y in R, either x >= y or x < y.
But R is not bounded (no inf, sup) and therefore has no minimum and no maximum.
Relatedly, every set gets a supremum and an infumum. (The first two must be members of the set. The second two can lie outside the set).
Without the extension, you need to make an exception for unbounded sets with the above rules.
> As there are infinitely many sets, instead of going through them manually, you have to find an algorithm [to select an item from each set]. So Person A tells the other player an algorithm to select an element. Then that other player finds a set where the algorithm fails.
> If you, for example say: "Take the smallest element", then I say "One of my sets is all negative integers, it doesn't have a smallest element."
> Then maybe you say: "Take the smallest element, or if it does not have that, take the largest." Then I say "One of my sets is all integers, which has neither". Then maybe you add: "If it has neither, take the element closest to 0 (in case of tie, take the larger one)". Then I say: "What about the set of all integers and all their reciprocals? It doesn't have an element closest to 0 either."
> When Person A gives up and says: "Look, there's some algorithm that does it, right, why do I have to recite it manually?", then that person has taken the Axiom of Choice. If they say "Okay, fine, you win, there's no algorithm you can't find a counterexample to", then that person rejects the Axiom of Choice.