1
$\begingroup$

So this a fundamental assumption in mathematics. Can someone explain informally what it actually is please.

My guess is that its when we say in proofs that "Let $x \in X$". But I am not sure.

$\endgroup$

marked as duplicate by Patrick Stevens, Henning Makholm, Charles, hardmath, Asaf Karagila axiom-of-choice Feb 3 '16 at 21:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ The body of a Question should be as self-contained as possible, not relying on the title to pose the essential problem. Please review How to Ask. $\endgroup$ – hardmath Feb 3 '16 at 21:53
0
$\begingroup$

Informally the axiom of choice says if you have a bunch of sets (possibly infinitely many) then you can choose one element from each set and build a new set out of them.

That you can do such a thing sounds at first incredibly intuitive and obvious. But over time mathematicians have realized that it is exactly that assumption that leads to a bunch of really disturbing results. Therefore some people have concluded that we shouldn't allow that assumption. However if you don't there are also a bunch of basic things that become much harder if not impossible to prove.

It's an ongoing debate with no end in sight.

$\endgroup$
  • $\begingroup$ so does this mean that you can just keep making sets after making new sets? So you can make infinite amount of sets? $\endgroup$ – snowman Feb 3 '16 at 21:47
  • $\begingroup$ Well sure you can always make an infinite amount of sets. But the AOC is saying something a bit more specific, it's saying from any collection of sets you can make a new set that has one element from each of them. It's that specific operation of choosing one element from each set that is involved in the AOC. $\endgroup$ – Gregory Grant Feb 3 '16 at 21:54
4
$\begingroup$

For the deepest results about partially ordered sets we need a new set-theoretic tool; ... We begin by observing that a set is either empty or it is not, and if it is not, then, by the definition of the empty set, there is an element in it. This remark can be generalized. If $X$ and $Y$ are sets, and if one of them is empty, then the Cartesian product $X\times Y$ is empty. If neither $X$ nor $Y$ is empty, then there is an element $x$ in $X$, and there is an element $y$ in $Y$; it follows that the ordered pair $(x,y)$ belongs to the Cartesian product $X\times Y$, so that $X\times Y$ is not empty. The preceding remarks constitute the cases $n=1$ and $n=2$ of the following assertion: if $\{ X_i\}$ is a finite sequence of sets, for $i$ in $n$, say, then a necessary and sufficient condition that their Cartesian product be empty is that at least one of them be empty. The assertion is easy to prove by induction on $n$. .... The generalization to infinite families of the non-trivial part of the assertion in the preceding paragraph (necessity) is the following important principle of set theory.

Axiom of choice: The Cartesian product of a non-empty family of non-empty sets is non-empty.

Paul Halmos Naive Set Theory page 59.

$\endgroup$
  • $\begingroup$ why downvotes? ${}$ $\endgroup$ – user153330 Feb 3 '16 at 21:36
  • 1
    $\begingroup$ I didn't downvote you, but I'm guessing it's because the OP asked for an "informal" explanation. $\endgroup$ – Gregory Grant Feb 3 '16 at 21:37
  • $\begingroup$ I didn't downvote $\endgroup$ – snowman Feb 3 '16 at 21:37
  • 2
    $\begingroup$ I'd like to register my opinion here. Unless a post is clearly vandalism or abusive then I think it's rude to downvote it without a constructive explanation. $\endgroup$ – Gregory Grant Feb 3 '16 at 21:39
  • $\begingroup$ @snowman do you want me to give you the entire motivation halmos gives so that the "informal" side of the axiom of choice becomes clear to you? $\endgroup$ – user153330 Feb 3 '16 at 21:41
0
$\begingroup$

Suppose you have a set $X$ (which may be infinite which is when things get exciting) and another set $Y$. Suppose that for each $x\in X$ you associate a non-empty subset $Y_x$ of $Y$, then there is a function $f:X\to Y$ such that $f(x)\in Y_x$. This seemingly intuitive "fact" (and in fact, trivial for finite $X$) is the Axiom of Choice. To be specific, the existence of $f$ is what the Axiom tells you. $f$ is a function that for each $x$ "chooses" in $f(x)\in Y_x$.

$\endgroup$
  • $\begingroup$ Ok, not very informal... think of having an infinite stall, with infinite amount of boxes, in each box there is a kind of fruit different from all the other boxes (and there are infinite kind of fruits). You can "choose" a shopping basket that has exactly one fruit per kind, and all kinds are accounted for. $\endgroup$ – Oskar Limka Feb 3 '16 at 21:43

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