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Suppose $F$ is an infinite set (that is $\#F\geq\#\mathbb N$). Various sources I have consulted claim that $$\# F=\# (F\times\mathbb N)$$ without proof (# denotes cardinality). I guess that this is so trivial that it should be immediate, yet I can't see it.

Obviously, $\# (F\times\mathbb N)\geq\#F$, because the function $(f,n)\mapsto f$ from $F\times\mathbb N$ to $F$ is clearly a surjection, but I can't prove the other direction, that $\#(F\times\mathbb N)\leq\#F$ (only in the particular case when $F$ is countable, i.e., $\#F=\#\mathbb N$). Once I could show this, the desired result would follow from the Schröder–Bernstein theorem.

Any help would be appreciated.

Asaf Karagila
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triple_sec
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    For infinite cardinals $\alpha,\beta$ --assuming AC-- we have $$\alpha\cdot\beta\ =\ \alpha+\beta\ =\ \max(\alpha,\beta),.$$ – Berci Feb 09 '14 at 23:41

1 Answers1

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You can't really prove this "outright", because the axiom of choice is necessary for establishing this result.

HINT: Assuming the axiom of choice, we have that $\#F=\#(F\times F)$, and that $F$ is infinite is the same as saying as $\#F\geq\#\Bbb N$.

Asaf Karagila
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  • I was just about to add: I have no reservations about using AC. – triple_sec Feb 09 '14 at 23:24
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    Of course. But the fact that it has to be used means that there's a good chance that you can't explicitly write down an injection. – Asaf Karagila Feb 09 '14 at 23:24
  • Do you know a reference for a simple proof of $#F=#(F\times F)$? – triple_sec Feb 09 '14 at 23:41
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    Well, if simple means "circumventing ordinals and transfinite induction" then not really. If you're okay with ordinals, or at least with transfinite recursion, then I have written several proofs on this site. – Asaf Karagila Feb 09 '14 at 23:44
  • I'm not very familiar with the theory of ordinal numbers, but I think I can deal with transfinite induction. Could you please provide a reference to such an answer of yours? – triple_sec Feb 09 '14 at 23:46
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    Well, proving $#F=#(F\times\Bbb N)$ is a bit easier then: using the axiom of choice you can get a transfinite sequence of all elements $F$, say $F={a_0,a_1,\dots,a_{\omega},a_{\omega+1},\dots,a_{2\omega},\dots,\dots}$. Since finite numbers don't matter for $F$, we can assume that this transfinite enumeration is divided by blocks of size $\omega:=|\Bbb N|$. Then replace each block to $\Bbb N\times\Bbb N$. – Berci Feb 09 '14 at 23:48
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    @triple_sec: http://math.stackexchange.com/questions/608538/godels-pairing-function-and-proving-c-cc-for-aleph-cardinals/608555#608555 – Asaf Karagila Feb 09 '14 at 23:48
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    @Berci: Yes, that also works. One can easily define an equivalence relation on $F$, $a_i\sim a_j$ if and only if both $[i,j]$ and $[j,i]$ are finite (one will be empty if $i\neq j$). This is a partition as wanted. – Asaf Karagila Feb 09 '14 at 23:49
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    @triple_sec: I sort of cut some corner there, but the idea as a whole should be clear. – Asaf Karagila Feb 09 '14 at 23:54
  • Guys, thank you so much for your help. I wound up consulting a Zorn-lemma-based proof (theorem 1.9 in this book) because I am more comfortable with this than with cardinals (yet the surprisingly annoying technical details almost made me give up understanding the detailed arguments even in the proof based on Zorn's lemma). Anyway, your input has been extremely useful. – triple_sec Feb 10 '14 at 03:39