1
$\begingroup$

Consider the sequence of sets $(A_1,A_2,A_3,\dots)$. Assume for each $A_n$, there exists bijective mapping $f_n: \{1,2,3,4\} \rightarrow A_n$ for each $n$.

If I want to construct the sequence $(f_1(2),f_2(2),f_3(2), f_4(2),\dots)$, do I need the Axiom of Choice (AC)? What if my statement above was changed to the following:

Consider the sequence of sets $(A_1,A_2,A_3,\dots)$ with a sequence of $(f_1,f_2,f_3,\dots)$ where each $f_n: \{1,2,3,4\} \rightarrow A_n$ for all $n$ and is bijective.

Motivation: Reading from set theory - Can you explain the “Axiom of choice” in simple terms? - Mathematics Stack Exchange, I learned that we need AC because we can not do an infinite number of existential instantiation in a a mathematical proof. So my current thought is that in the first statement, since we need to use an infinite number of existential instantiation, one for each instantiation of the function, we need AC; while in the second statement, I am already given that sequence of function, so there is no need for AC.

It will be great if someone can illustrate this concept with some more concrete examples like when AC is needed and when it is not.

$\endgroup$
  • $\begingroup$ For both you don't need AC. $\endgroup$ – Kenny Lau Oct 3 '17 at 23:04
  • $\begingroup$ I take that back. Apparently you did not assume "there exists a unique bijecive mapping". You would need AC then. $\endgroup$ – Kenny Lau Oct 3 '17 at 23:15
  • $\begingroup$ I don't understand your question. Are the $f_n$ given? If so, then how are they given? If not, then what would $f_1(2)$ mean? $\endgroup$ – Kenny Lau Oct 3 '17 at 23:16
  • $\begingroup$ @KennyLau I just updated the question slightly to say there exist an $f_n$ for each $n\in\mathbb{N}$. Why would I not need AC if I assume uniqueness? In the first statement, $f_n$ are not given. I simply stated they exist for each $n\in\mathbb{N}$. I am not sure how they will be given because I don't know if AC is needed for the sequence $(f_1,f_2,f_3,\dots)$ to be given, and that is part of the question. $\endgroup$ – Kun Oct 3 '17 at 23:24
  • $\begingroup$ Because if you assume uniqueness, you can apply the axiom schema of replacement to get the sequence $(f_1,f_2,f_3,\cdots)$. $\endgroup$ – Kenny Lau Oct 3 '17 at 23:25
2
$\begingroup$

The answer to "when do you need choice" is a bit complicated. There are many mitigating factors. It's like asking in simple terms when do you need to add flour to your sauce. There is a rule of thumb, but there are a lot of variations which make this rule of thumb only a rough guideline.

The simplistic answer is that you don't need the axiom of choice if there is a concrete rule that you can write which specifies a unique element from each set.

But what does it even mean? Well. Recall that when writing rules we are allowed to use some parameters. If you are choosing from a sequence of groups (which are sets with a given group structure), then you can always choose the identity element, for example. If you are choosing from a collection of sets of natural numbers, you can always choose the minimal, or the least odd if there is one, or the third even, etc. There are many ways to define this.

So when you are given a sequence of bijections between your sets and $\{0,1,2,3\}$, or any fixed set for that matter, choosing is easy, since you can just choose all the preimages of a fixed element.

The problem begins when you don't have a way to specify these bijections, and you can just prove that for each and every set in your collection there is a bijection. This usually means there are many bijections. So how do you choose one for each set? This is again a problem of choice. If there is some mitigating factors, if there is a way to uniformly define a bijection for each set, we're good. But otherwise, choice is needed.

For example, there is a unique order preserving bijection between any set of natural numbers and an initial segment of the natural numbers, even if there are many bijection, there is a special one. So given a sequence of sets of natural numbers, we can look at the sequence of order-preserving bijections. For arbitrary sets, this is not necessarily true anymore; as demonstrated by Russell in his famous "socks" example, and formalized by Fraenkel and later by Cohen in rigorous set theoretic terms.

To sum up, when you can give a uniform way of choosing an element from each set, you do not need the axiom of choice. This can be in the form of bijections with a fixed set, or some given structure which allows you to distinguish an element in a uniform way.

However, just because each of your sets has a bijection, or a group structure, or whatever, does not mean that you can uniformly fix one for all of them. That means that you choose from the family of "relevant structures", which is itself a use of the axiom of choice.


All this assumes that the family of sets is infinite. Of course, if it is finite, then we can prove using induction that we can always choose.

$\endgroup$
  • $\begingroup$ So basically when we can give a uniform way of choosing element, we can invoke the axiom schema of replacement and get the desired collection of elements? $\endgroup$ – Kun Oct 4 '17 at 9:49
  • $\begingroup$ Yes. That is exactly the point. It is important to remark that "uniform" allows parameters, which is something people often tend to overlook (formally, but intuitively they usually think "with parameters" anyway). $\endgroup$ – Asaf Karagila Oct 4 '17 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.