User talk:Haran

Your question on the talk page of the "axiom of choice" article

I know very little about set theory, and even less about Wikipedia, but hopefully I'll be of use nonetheless.

The answer in the link is actually pretty clear, and is true—though is might not be written well enough to use as a source. Given that you've read a little on set theory, I might be able to explain some of it to you. Here we go:

First, let's try to understand what "the number of parts does not exceed the number of elements of the set being partitioned" means. Because set theory isn't restricted to finite sets, the terms "number" and "does not exceed" have to be interpreted differently than usual. In most cases they're interpreted separately, with "number of (elements of)" interpreted as cardinality and "does not exceed" interpreted as referring to the order of cardinals. However, in contexts where cardinals/cardinality is not defined (which include all contexts where the axiom of choice isn't a given) there's another way to say a set $A$ isn't bigger than a set $B$—that's true if there's an injective function $f$ from $A$ to $B$. That, by the way, is true because if there's such a function, then surely there are exactly as many elements in $A$ as in the image of $f$ (which is contained in $B$) and thus $B$ has at least as many elements as $A$.

Now, to the proof.

Say the axiom of choice is true. Why is it always true that when you partition a set into disjoint nonempty parts, "the number of parts does not exceed the number of elements of the set being partitioned"? Well, that means if you have a set $B$ and you partition it into disjoint nonempty parts, and denote the set of those parts by $A$, then "the number of elements of $A$ does not exceed the number of elements of $B$", that is, there's an injective function $f$ from $A$ to $B$. So let's indeed prove that.

By the axiom of choice, there's a choice function $f$ defined on $A$. It's injective because the partition splits $B$ into disjoint subsets, and $\bigcup A=B$, so $f$ is from $A$ to $B$.

That was it. Quick and simple, except the introduction is a nightmare. However, it only explained some of the claim in that link: I didn't prove that "this can fail without the axiom of choice". What that means is "without using the axiom of choice, this can't be proven". I don't know how to prove that one, though forcing (which I know nothing about) is probably related to that.

If you reply to me, please notify me by linking to my user page ([[User:Professor Proof]] or [[User:Professor Proof|Professor Proof]] would suffice) somewhere in your message and not forgetting to sign (you have to link and sign in the same edit or it won't work).

If you have any questions on this or other matters, feel free to ask! Professor Proof (talk) 02:44, 1 June 2018 (UTC)