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It happens a lot to me that when I find an intuitive model (picture) of a mathematical entity, the proofs left as exercises in books are very easy to solve. For example when dealing with filters and ultrafilters on sets (specially $\omega$) I just need to imagine the Hasse diagram of the Poset $\langle\mathcal{P}(\omega),\subseteq\rangle$ and most proofs and definitions come naturally.

I have been trying to find a model/picture for arithmetic with cardinals that helps me solve the problems that come in some books but I don't know if cardinal numbers are too big and I cannot picture them correctly or I just have to use different models/pictures for different problems. So far the two that have been working best are parallel lines (when comparing cardinals) and my intuition with injective and surjective functions. But those approaches only work quickly with easy problems (adding or multiplying finitely many cardinals or relating the cardinality of two specific sets). However, this techniques become lengthy when new concepts are introduced (like cofinality and exponentiation of cardinals). Moreover, I have not been able to solve many problems with just these two methods. I'll quote two of those problems and try to make this ideas more clear:

  1. (This one is in Andras Hajnal & Peter Hamburguer's Set Theory Book) If $\kappa$ is an infinite cardinal number and $\kappa=\sum_{\lambda<cf(\kappa)}\kappa_\lambda$ for some sequence $\langle\kappa_\lambda\rangle_{\lambda<cf(\kappa)}$ of cardinals then we have $\kappa^{cf(\kappa)}=\prod_{\lambda<cf(\kappa)}\kappa_\lambda$.

For this one I just changed $\kappa^{cf(\kappa)}$ for $\prod_{\lambda<cf(\kappa)}\kappa$. Then you can easily get that $\prod_{\lambda<cf(\kappa)}\kappa_\lambda\leq\prod_{\lambda<cf(\kappa)}\kappa$. However, for the nontrivial inequality: $\prod_{\lambda<cf(\kappa)}\kappa\leq\prod_{\lambda<cf(\kappa)}\kappa_\lambda$, searching an injective function doesn't seem like the best way to go and I believe that with a better picture the argument would just come to me in an "arithmetical" way (and not with functions).

  1. (A friend gave me this one but I have a feeling there is a typo somewhere) Suppose that $\alpha$ is a limit ordinal and that $\langle\kappa_\xi\rangle_{\xi<\alpha}$ is a strictly increasing sequence of cardinals such that $\kappa=\sum_{\xi<\alpha}\kappa_\xi$ and if $0<\lambda<cf(\alpha)$ then $\kappa^\lambda=\sum_{\xi<cf(\kappa)}\kappa^\lambda_\xi$.

Although not stated in the problem, every cardinal $\kappa_\xi$ must be strictly less than $\kappa$ (otherwise the sum is greater than $\kappa$). And given that for every infinite cardinal $cf(\kappa)=\min\{\lambda\in CN_\infty\mid\forall\xi\in\lambda(\kappa_\xi<\kappa)\wedge\kappa=\sum_{\xi<\lambda}\kappa_\xi\}$ we must have that $cf(\kappa)\leq\alpha$. Under the same principle one just have to prove that $cf(\kappa^\lambda)\leq\alpha$. There is also a trivial inequality in this problem: $\kappa^\lambda=(\sum_{\xi<\alpha}\kappa_\xi)^\lambda\geq\sum_{\xi<cf(\kappa)}\kappa^\lambda_\xi$. But those are just syntactical conclusions from the premises and I could not think of a mental image to solve the problem.

So the questions are: Do you use a single model/picture to help you solve this kind of problems about infinite sums, products and exponentiation of cardinals? Which one? and could you provide a hint on how to use such a model/picture to solve problems 1. and 2.?

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  • $\begingroup$ You should use \sum and \prod for $\sum$ and $\prod$. Not \Sigma and \Pi. $\endgroup$ – Asaf Karagila Mar 17 '15 at 6:21
  • $\begingroup$ @AsafKaragila yep, sorry, in the rush I just picked the first names that came to my mind. Changed all of them... $\endgroup$ – Jonathan Julián Huerta Mar 17 '15 at 6:36
  • $\begingroup$ Do you have a way to visualize products over an infinite index set? $\endgroup$ – Brian M. Scott Mar 17 '15 at 10:07
  • $\begingroup$ @BrianM.Scott I usually visualize its elements as "infinite uncountable tuples/sequences" (if that makes sense) however, as cardinal numbers I have not a clear visualization of them. $\endgroup$ – Jonathan Julián Huerta Mar 17 '15 at 15:58
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I just realized that you are asking for other methods than constructing explicit functions, however, I figure direct attacks might be the most intuitive for problems like this.

In order to show that $\kappa^{cf(\kappa)}\leq\prod_{\lambda<cf(\kappa)}\kappa_\lambda$ it is helpful to visualize the LHS as collection of functions from $cf(\kappa)$ to $\kappa$ and RHS as a sequence $\langle a_\lambda : \lambda\in cf(\kappa) \rangle$ such that $a_\lambda\in \kappa_\lambda$. Constructing injections from one to the other seems the most natural to me so you could do the following:

Given $f\in \kappa^{cf(\kappa)}$, recursively build a sequence $g$ such that $g(\lambda)=\begin{cases} f(i)+1 && \text{if $i$ is the least such that f(i) is in }[0,\kappa_\lambda) \text{ and i is not used previously} \\ 0 && otherwise \end{cases} $.

Note we eventually exhaust all f-values as $cf(\kappa)$ is regular.

If $f\neq h$, then let $i\in cf(\kappa)$ be the least point where they differ. There are two cases to consider one is there exists a least undefined $\lambda$ such that $f(i), h(i)\in [0,\kappa_\lambda)$ and another case being there exists a least undefined $\lambda$ such that $\kappa_\lambda$ separates $f(i)$ and $h(i)$. In either case, the sequences constructed from two functions are different. So you have an injection.

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  • $\begingroup$ Ok, thank you very much, I like the approach you took for this one. I'll try to do the same with the other exercises. Do you think/feel that's then the right intuition? A direct attack seeing the Cardinal Numbers as Sets? $\endgroup$ – Jonathan Julián Huerta Mar 20 '15 at 3:33
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    $\begingroup$ Yeah. At least that is how those cardinals are defined, as some well-ordering. $\endgroup$ – Jing Zhang Mar 20 '15 at 3:36

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