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Let $A$ and $B$ be two sets. Prove the existence of an injection from $A$ to $B$ or an injection from $B$ to $A$.

I don't know how to proceed, since I don't have any information on $A$ or $B$ to begin with.

Does anybody have a hint ?

Cameron Buie
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Gabriel Romon
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    Have you been studying cardinality? – Brian M. Scott Oct 31 '13 at 17:04
  • I know the basics about it. – Gabriel Romon Oct 31 '13 at 17:05
  • Either $|A|\le|B|$, or $|B|\le|A|$; what does that say in terms of injections? – Brian M. Scott Oct 31 '13 at 17:06
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    Yeah, this isn't trivial. People usually use the well ordering theorem for that. – Dan Shved Oct 31 '13 at 17:07
  • @Brian Well how to prove your statement : either $|B|\le|A|$ or $|A|\le|B|$ ? – Gabriel Romon Oct 31 '13 at 17:08
  • Depends on what you’ve been taught about cardinality. If this is an elementary course, I’d expect you to have been taught that for any two sets $A$ and $B$, exactly one of $|A|<|B|$, $|A|=|B|$, and $|B|<|A|$ holds. This is a form of the axiom of choice, and at an elementary level it’s generally assumed without much (if any) discussion. – Brian M. Scott Oct 31 '13 at 17:09
  • I've not been taught this unfortunately. Is it this hard to prove ? – Gabriel Romon Oct 31 '13 at 17:11
  • @GabrielR. The axiom of choice cannot be proved... – AlexR Oct 31 '13 at 17:12
  • OK. Can anybody give a link to the proof of the previous statement ? – Gabriel Romon Oct 31 '13 at 17:14
  • @Gabriel: Have you been taught anything about either the axiom of choice or the well-ordering theorem? Because if not, I suspect that you’re supposed to assume that $|B|\le|A|$ or $|A|\le|B|$ for any sets $A$ and $B$. – Brian M. Scott Oct 31 '13 at 17:15
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    You can also use Zorn's Lemma - pick a maximal bijection between subsets. Of course Zorn is nothing else but yet another varaiant of teh Axiom of Choice. – Hagen von Eitzen Oct 31 '13 at 17:17
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    Nope. But it's not a problem for me to start from scratch, even if it involves learning about the axiom of choice. – Gabriel Romon Oct 31 '13 at 17:20
  • @GabrielR. I personally liked this post in Terence Tao's blog that explains how to derive things like Zorn's lemma and the well-ordering theorem from the Axiom of Choice, and also explains what exactly is meant by the cardinality of a set. Although others can probably give you links to more standard sources. – Dan Shved Oct 31 '13 at 17:24
  • More than a handful of previous discussions on the site: (1) http://math.stackexchange.com/q/182386/622 (2) http://math.stackexchange.com/q/268942/622 (3) http://math.stackexchange.com/q/69774/622 (4) http://math.stackexchange.com/q/151904/622. Some discuss the necessity of the axiom of choice; others discuss the proof from the axiom of choice. – Asaf Karagila Oct 31 '13 at 17:40

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To prove the existence of an injection between two sets $A$ and $B$ from the Axiom of Choice, or from Zorn's Lemma (which is equivalent) the rough idea is that we build a bijection between larger and larger subsets of $A$ and $B$, starting with the empty function. At every "step" we match one of the remaining elements of $A$ to one of the remaining elements of $B$, continuing until we run out of elements in one (or both) of the sets. When we run out of elements in one of the sets, say in $A$ first, it means that we have matched every element of $A$ to an element of $B$, giving an injection $A \to B$. If instead we run out of room in $B$ first, we get an injection $B \to A$.

There are two different ways to formalize this idea. The first one uses the Axiom of Choice to choose, at every step, elements of $A$ and $B$ from among those which haven't yet been paired up, so that we can pair them up and extend our function. Once we have a pair of "choice functions" $f$ and $g$ that choose elements of $A$ and $B$ for us respectively, the notion of proceeding by infinitely many "steps" is formalized using transfinite recursion.

Instead of using the combination of the Axiom of Choice and transfinite recursion, one can use Zorn's Lemma instead. This works out to be essentially the same argument overall, but it has the advantage that it abstracts away the condition you need to check in order to apply transfinite recursion, so can just check this condition and you don't need to know how transfinite recursion works.

Trevor Wilson
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