Timeline for Why worry about the axiom of choice?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2023 at 13:14 | comment | added | Timothy Chow | As Jerry Bona said, “The axiom of choice is obviously true, the well-ordering theorem is obviously false; and who can tell about Zorn’s lemma?” | |
| Aug 17, 2019 at 20:24 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Jan 23, 2018 at 18:22 | comment | added | Anton Fetisov | It isn't natural to believe that a product of nonempty sets is nonempty once you generalize a bit: an $I$-indexed family of sets $\{ J_i \}_{i\in I}$ is the same as an epimorphism $J \to I$, $J = \sum_{i\in I} J_i$. An element of a product of this family is the same as a section of this epimorphism --- and of course epimorphisms in categories can have no sections! This is true even in categories that are a model of (extensional) set theory, i.e. in toposes like a category of sheaves of sets on a space. E.g. for any nontrivial manifold $X$ its $Sh(X)$ will not satisfy AC. | |
| Oct 16, 2013 at 1:32 | comment | added | Kaveh | The two statements are very close to each other so it might worth to explore why we have mentally different intuitions about the two statements. I think it might have to do with the word "choice" as if we need some action performed by an agent while for the other one there is no such word. | |
| Oct 10, 2013 at 17:21 | comment | added | Peter LeFanu Lumsdaine | @NeilToronto: “every set has a unique cardinality” can be phrased in a lot of different ways, plenty of which are provable without choice (e.g.: there is a class $\mathbf{Card}$ and a “cardinality” map $V \to \mathbf{Card}$, such that sets have the same cardinality iff they are isomorphic). The only versions of the statement I know that are equivalent to AC are ones which insist that each cardinality should be represented by some ordinal — but this is a very thinly veiled version of the well-ordering principle, and I think not at all so intuitively obvious. | |
| Oct 10, 2013 at 14:59 | comment | added | Neil Toronto | Another fine equivalence to AC: every set has a unique cardinality. This and your first example are the main things that convince me AC is not so strange. | |
| Oct 8, 2013 at 23:13 | comment | added | Qfwfq | If you construct $\mathbb{C}$ as $\mathbb{R}^2$ with product $(a,b)\cdot (c,d):=(ac-bd,bc+ad)$, then $i:=(0,1)$ is a standard definition. Also if you construct $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$ you have a standard choice: $i:=x\mathrm{mod}(x^2+1)$. | |
| Aug 16, 2013 at 1:26 | comment | added | Vladimir Reshetnikov | Wait, the one that's above the real line is $i$, and the one below is $-i$, right? ;) | |
| May 2, 2010 at 1:06 | history | answered | Daniel Asimov | CC BY-SA 2.5 |