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If $ \in$ Card where Card is the class of all infinite cardinals, and if $0 \neq \lambda \in$ Cd where Cd is the class of all cardinals, how can I prove the followings:

(but without possible usage of Godel's pairing function)

$a)$ $\kappa \times \kappa = \kappa$

$b)$ $\kappa \times \lambda =$ max {$\kappa , \lambda$}

$c)$ $\kappa + \lambda =$ max {$\kappa , \lambda$}

Remark: I use the following definition of cardinals:

card$(x):=$ min {$\alpha \in$ Ord | $\exists$ bijection $f:\alpha \to x$}

edward_scissorhands
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1 Answers1

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We'll first assume that we proved a) and deduce b) and c). Then we'll prove a) (obviously without using our newly acquired results).

So let $\kappa, \lambda$ be two infinite cardinals. Assume $\lambda\leq \kappa$. We let $x\leq y$ denote "there is an injection $x\to y$". We have $\kappa \leq \kappa +\lambda \leq \kappa\times\lambda \leq \kappa\times\kappa\leq \kappa$; which allows us to conclude for b) and c) using Cantor-Bernstein's theorem. The first inequality is trivial, the second one is obtained by $\kappa +\lambda \sim \kappa\times\{0\}\cup\lambda\times\{1\}$ and then using the inclusion map (here we use $\lambda\neq 0$); the third one comes from our assumption that $\lambda\leq \kappa$, and the final one comes from a). In other words, it suffices to prove a).

The proof for a) is a bit more delicate, as it uses induction, but I'm sure you must be familiar with this. The result is obviously true for $\omega$ (there are many explicit bijections $\omega\times\omega\to\omega$). Therefore let $\kappa$ be any infinite cardinal and assume the result holds for all cardinals $<\kappa$.

Define the following order on $\kappa\times\kappa$: $(x,y)\preceq (x', y')$ if and only if : either $max(x,y) < max(x',y') $ or ($max(x,y)=max(x',y')$ and $(x,y)\leq (x', y')$ ) (in the last condition, $\leq$ denotes the lexicographic ordering). It's quite easy to see that this is a well ordering on $\kappa\times\kappa$ (I'll leave the details to you)

Therefore, $(\kappa\times\kappa,\preceq)$ is isomorphic (with a unique isomorphism $f$) to a unique ordinal $\delta$. What we want to show is that $\delta$ cannot be bigger than $\kappa$, which will then show that it is $\kappa$.

So assume $\kappa < \delta$. Then let $(\lambda_0, \lambda_1)= f^{-1}(\kappa)$. Both $\lambda_i <\kappa$, so if we let $\lambda = sup(\lambda_0,\lambda_1)$, we have $\lambda<\kappa$.

Since $f$ is an isomorphism, $\kappa \subset Im(f_{\mid \lambda\times \lambda})$. But using our induction hypothesis, $\lambda\times\lambda$ has a cardinal $<\kappa$ : indeed it is either finite, in which case it's obvious, or it is infinite and then its cardinal is the cardinal of $\lambda <\kappa$. This is a contradiction, since $Im(f_{\mid \lambda \times \lambda})$ has cardinal at most $card(\lambda\times\lambda)$.

Therefore our assumption was wrong, $\delta\leq \kappa$, and so $\kappa\times\kappa \leq \kappa$. The reverse inequality being trivial, Cantor-Bernstein's theorem shows that $\kappa\times \kappa \sim\kappa$.

This gives us a), and thus b) and c)

Maxime Ramzi
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  • Isn't your order on $\kappa\times\kappa$ just the Gödel pairing that OP explicitely wants to avoid? – M. Winter Feb 21 '17 at 09:26
  • I don't get how $f^{-1}(\kappa)$ can be $(\lambda_0,\lambda_1)$. I mean $f:\kappa\times\kappa\rightarrow\delta$ is an isomorphism between well-ordererings and $\kappa$ is an inital segment of $\delta$. So $f^{-1}(\kappa)$ must be an initail segment of $\kappa\times\kappa$. How can this possibly be a pair $(\lambda_0,\lambda_1)$? Can you please explain? – M. Winter Feb 21 '17 at 09:36
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    Oh I hadn't noticed that he wanted to avoid using Gödel pairing... well he seems to have validated my answer so maybe it doesn't bother him ? As for $f^{-1}(\kappa)$, I wasn't considering $\kappa$ as an initial segment, but as an element : $f^{-1}$ isn't the inverse image, it's the unique couple $(x,y)$ such that $f(x,y)=\kappa$ – Maxime Ramzi Feb 21 '17 at 11:06
  • Ah I see! I understand it now. Thank you. – M. Winter Feb 21 '17 at 11:12
  • Where is the axiom of choice used in the proof? – Vivaan Daga Jul 22 '22 at 01:22
  • @Shinrin-Yoku : it isn't - the statement is true in ZF (crucially it's using the definition of cardinal from the end of the question) – Maxime Ramzi Jul 22 '22 at 06:45
  • Thanks, so why can't we use a transfinite induction style argument starting with $\aleph_{0}$ and at each step unioning a set of the same cardinality? – Vivaan Daga Mar 28 '23 at 17:06
  • @Shinrin-Yoku : the proof does work by induction :D The problem of trying to "add something" by hand is simply that to go from, say $A\times A$ to $B\times B$, you need to add $(B\setminus A)\times (B\setminus A) \cup (B\setminus A)\times A \cup A\times (B\setminus A)$, and the bijection with $(B\setminus A)$ becomes hard to set up. – Maxime Ramzi Mar 28 '23 at 19:39
  • But we can establish $\aleph_{k}+\aleph_{k}=\aleph_{k}$ by using a similar “doing by hand” recursion argument. Of course the whole point of using alephs is to avoid choice. – Vivaan Daga Mar 28 '23 at 19:54
  • @Shinrin-Yoku : it's not clear to me. The only argument I can come up with for sums is essentially a variant of the one I wrote in my answer - if you have a simpler argument in mind, it would be very interesting to see : – Maxime Ramzi Mar 28 '23 at 19:57
  • It’s a similar recursive argument you start with $\aleph_{0}$ and at each stage this time you union a set of cardinality $\aleph_{0}$. – Vivaan Daga Mar 29 '23 at 03:13
  • @Shinrin-Yoku : Again, I don't see how that would work, but if you do, it could be helpful to write an answer :) – Maxime Ramzi Mar 29 '23 at 09:07
  • Say we have an aleph, $\aleph_{k}$, to prove that $\aleph_{k}+\aleph_{k}=\aleph{k}$ we proceed as follows, start with $\aleph_{0}$ a subset of $\aleph_{k}$, clearly we have the existence of a bijection $f$ from $\aleph_{0} \to 2\aleph_{0}$, now we start a transfinite process, by unioning a set of cardinality $\aleph_{0}$ from $\aleph_{k}\setminus \aleph_{0}$(the set can be chosen canonically so no choice required), we can show that for this new set, X, formed by the union there is a bijection $g$ extending $f$ such that $g$ is a bijection from X to 2X like this we continue the transfinitely – Vivaan Daga Mar 29 '23 at 09:19
  • And can show that $\aleph_{k}$ has the desired property – Vivaan Daga Mar 29 '23 at 09:19
  • Does that clarify things? – Vivaan Daga Mar 29 '23 at 09:20
  • @Shinrin-Yoku : yeah, ok, I guess for sums it works, sorry, I was still worrying about products. But if you unravel it, it does exactly what you want it to, namely it sets up "the canonical well-ordering on $\kappa + \kappa$" , and then this sort of tells you what to do for the product as well. – Maxime Ramzi Mar 29 '23 at 09:46
  • Another interesting question suppose we use this pairing function https://en.m.wikipedia.org/wiki/Pairing_function#/media/File%3ACantor's_Pairing_Function.svg, is the order type of $\aleph_{\alpha}\times \aleph_{\alpha}$ equal to $\aleph_{\alpha}$? – Vivaan Daga Mar 30 '23 at 17:00
  • Any ideas for that? – Vivaan Daga Mar 31 '23 at 04:22
  • @Shinrin-Yoku : how exactly do you generalize this pairing function to general $\aleph_\alpha$ ? Do you use ordinal sum ? There might be issues in doing so related to the lack of continuity of $\alpha \mapsto \alpha + \beta$. – Maxime Ramzi Mar 31 '23 at 08:45
  • Good point..... – Vivaan Daga Mar 31 '23 at 14:28