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Let $Card$ be the class of infinite cardinals and $p\colon Card^2\to Card$ be given by $(\kappa,\lambda)\mapsto\kappa^\lambda$.
Assuming GCH it is known that $p(\kappa,\lambda)$ is either $\kappa$ (if $\lambda < cof(\kappa)$) or $\kappa^+$ (if $cof(\kappa)≤\lambda ≤\kappa$) or $\lambda^+$ (if $\kappa < \lambda$).
However, knowing less can be enough. For instance the function $\gimel\colon \kappa\mapsto\kappa^{cof(\kappa)}$ completely determines $p$.

Generalizing from powers to arbitrary products yields the following natural question:

Consider a sequence $(\kappa_i\mid i<\delta)$ of (infinite) cardinals. Can the value $\prod\limits_{i<\delta}\kappa_i$ be de determined knowing (all values of) $\gimel$ or assuming GCH?

One can reformulate the question as: under which circumstances is $\prod\limits_{i<\delta}\kappa_i$ determined by $p$?

EDIT: So I guess this is the final result:
For $(\kappa(i)\mid i<\delta)$ nondecreasing sequence of infinite cardinals we have for $\delta=\lambda_0\alpha_0+\ldots+\lambda_l\alpha_l$ (every ordinal can be written this way) and $\delta_k:=\lambda_0\alpha_0+\ldots+\lambda_{k}\alpha_{k}$ with cardinals $\lambda_0>\ldots>\lambda_l$ and ordinals $0<\alpha_k<\lambda_k^+$:
$\prod\limits_{i<\delta}\kappa(i) = \max\limits_{0 ≤ k ≤ l} \left(\sup\limits_{i<\lambda_k\alpha_k}\kappa(\delta_{k-1}+i)\right)^{\lambda_k}$

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I'm fairly certain that GCH determines pretty much everything. Because everything is as small as it can be. –  Asaf Karagila Feb 22 '13 at 0:46
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The product of infinite cardinals is determined by $p$ via: $\prod_{i<\delta} \kappa_i = (sup \kappa_i)^{\delta}$. –  Stefan Hoffelner Feb 22 '13 at 9:49
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that doesn't seem right to me. assuming $\aleph_0^{\aleph_0}<\aleph_\omega$ (which is certainly possible) and taking $\kappa_i:=\aleph_0$ for $i<\omega$ and $\kappa_\omega:=\aleph_\omega$ we get $\prod_{i<\omega+1}\kappa_i = \aleph_0^{\aleph_0}\cdot\aleph_\omega=\aleph_\omega$ but $(\sup \kappa_i)^{|\delta|}=\aleph_\omega^{\aleph_0}=\gimel(\aleph_\omega)>\aleph_\omeg‌​a$ (since $\aleph_0=cof(\aleph_\omega)$) –  Toink Feb 22 '13 at 11:57
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Indeed, what Stefan suggested seems to only hold for cardinal $\delta$ and nondecreasing $\kappa_i$. –  Miha Habič Feb 22 '13 at 22:15
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Stefan's formula can't hold for arbitrary limit $\delta$: consider $\delta=\omega_1+\omega$. You can construct a counterexample similar in spirit to Toink's for $\omega+1$ (even one in which the $\kappa_i$ are strictly increasing). The argument I use at the end of my answer shows that it holds when $\delta$ is a cardinal and the $\kappa_i$ are nondecreasing. –  Eric Wofsey Feb 22 '13 at 23:43

4 Answers 4

Arbitrary products are determined by exponentiation even without GCH. As in the second part of Toink's answer, I assume the $\kappa_i$ are nondecreasing and unbounded with limit $\kappa$.

First, note that $cf(\delta)=cf(\kappa)$. Choose a cofinal sequence $(i_\alpha)$ in $\delta$ of length $\gamma=cf(\kappa)$. I claim it is possible to choose this sequence such that $f(\alpha)=|[i_\alpha,i_{\alpha+1})|$ is eventually a nondecreasing function of $\alpha$. First, if $f(\alpha)<\gamma$ for all sufficiently large $\alpha$, by regularity of $\gamma$ we can replace the $i_\alpha$ with a subsequence for which $f(\alpha)$ is either eventually always the predecessor of $\gamma$ (if $\gamma$ is successor) or eventually an increasing sequence of cardinals approaching $\gamma$ (if $\gamma$ is limit).

If $f(\alpha)\geq \gamma$ for arbitrarily large $\alpha$, let $\eta=\limsup f(\alpha)$. If $f(\alpha)=\eta$ for arbitrarily large $\alpha$, we can choose a subsequence of the $i_\alpha$ such that $f(\alpha)=\eta$ for all sufficiently large $\alpha$. Otherwise, if $cf(\eta)<\gamma$, we can similarly find a subsequence to make $f(\alpha)=\eta$ for all sufficiently large $\alpha$ (since $f(\alpha)$ must get arbitrarily close to $\eta$ over and over again as $\alpha$ approaches $\gamma$). Finally, if $cf(\eta)=\gamma$, we can pick a subsequence such that $f(\alpha)$ is eventually an increasing sequence of cardinals approaching $\eta$.

Now $$\prod \kappa_i=\prod_{\alpha<\gamma} \left(\prod_{i_\alpha}^{i_{\alpha+1}} \kappa_{\beta}\right),$$ and by induction on $\delta$ we can compute each of these smaller products. By our choice of the sequence $i_\alpha$, these smaller products are also eventually nondecreasing; splitting off an initial segment and again using induction, we may assume they are actually nondecreasing. As long as they are not eventually constant, the cofinality of their supremum will be $\gamma$. Thus we have reduced the problem to the case $\delta=\gamma=cf(\kappa)$.

Now by choosing a bijection $\delta=\delta^2$ we can split the product into $\delta$ many products, each of size $\delta$. Each one of these subproducts must be at least $\sup \kappa_i=\kappa$, so the entire product is at least $\kappa^\delta$. But the product is also clearly at most $\kappa^\delta$, so we have $\prod \kappa_i=\kappa^\delta$.

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Here I show how GCH determines every (infinite) product of infinite cardinals. I still don't know about the general question though.

Without loss of generality we have $\kappa_i ≤ \kappa_j$ for $i< j$.

First assume that there is some $\gamma <\delta$ such that $\kappa_\gamma$ is maximal. Then we have $\prod\limits_{i< \delta}\kappa_i=\left(\prod\limits_{i< \gamma}\kappa_i\right)\cdot\kappa_\gamma^{|\delta-\gamma|}$ where $\delta-\gamma$ is the unique ordinal $\epsilon$ such that $\gamma + \epsilon = \delta$. The first factor can be considered to be determined by induction on $\delta$ so we are done here.

So we can assume that there no maximum and set $\kappa := \sup\limits_{i<\delta}\kappa_i$, which means that for all $i<\delta$ we have $\kappa_i<\kappa$. Also observe that $\delta ≥ cof(\kappa)$ and that $\delta$ is a limit ordinal.

From now on assume GCH.

If $|\delta|≥\kappa$ we have $|\delta|^+=\aleph_0^{|\delta|}≤\prod\limits_{i<\delta}\kappa_i≤\kappa^{|\delta|}=|\delta|^+$, so assume $|\delta|<\kappa$ from now on.

If we assume $\kappa_i<\kappa_{i+1}$ for all $i<\delta$ we get by König: $\kappa=\bigcup\limits_{i<\delta}\kappa_i≤\sum\limits_{i<\delta}\kappa_i<\prod\limits_{i<\delta}\kappa_{i+1}≤\kappa^{|\delta|}=\kappa^+$, which implies $\prod\limits_{i<\delta}\kappa_i=\prod\limits_{i<\delta}\kappa_{i+1}=\kappa^+$

Now write $\prod\limits_{i<\delta}\kappa_i=\prod\limits_{i<\gamma}\kappa_{\alpha(i)}^{\epsilon(i)}$ for appropriate $\gamma≤\delta<\kappa$, $\alpha$ strictly increasing such that $\kappa_{\alpha(i)}<\kappa_{\alpha(j)}$ for $i< j$ and appropriate $\epsilon$. Note that $(\kappa_{\alpha(i)}\mid i<\gamma)$ is still cofinal in $\kappa$, so $\gamma≥cof(\kappa)$.

Therefore we get: $\kappa^+≥\prod\limits_{i<\gamma}\kappa_{\alpha(i)}^{\epsilon(i)}≥\prod\limits_{i<\gamma}\kappa_{\alpha(i)}=\kappa^+$ where the last equality holds by the previous results replacing $\delta$ by $\gamma$ and $\kappa_i$ by $\kappa_{\alpha(i)}$.

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up vote 1 down vote accepted

I guess this is the final result:
For $(\kappa(i)\mid i<\delta)$ nondecreasing sequence of infinite cardinals we have for $\delta=\lambda_0\alpha_0+\ldots+\lambda_l\alpha_l$ (every ordinal can be written this way) and $\delta_k:=\lambda_0\alpha_0+\ldots+\lambda_{k}\alpha_{k}$ with cardinals $\lambda_0>\ldots>\lambda_l$ and ordinals $0<\alpha_k<\lambda_k^+$:
$\prod\limits_{i<\delta}\kappa(i) = \max\limits_{0 ≤ k ≤ l} \left(\sup\limits_{i<\lambda_k\alpha_k}\kappa(\delta_{k-1}+i)\right)^{\lambda_k}$


Lemma 1: Each limit ordinal $\alpha$ can be written as $\sum\limits_{i< l}\lambda_i\alpha_i$ where $(\lambda_i\mid i< l)$ is a strictly decreasing finite sequence of infinite cardinals and $(\alpha_i|i< l)$ are some ordinals with $\alpha_i<\lambda_i^+$.

Proof: Take some $\alpha\in Lim$. Let $\lambda_0:=|\alpha|$ and $\beta:=min(\gamma\in Ord\mid \lambda \cdot \gamma >\alpha)$. If $\beta$ was a limit we would have $\lambda\beta = \sup\limits_{\gamma<\beta}\lambda\gamma≤\alpha$, so $\beta$ is a successor and we can write $\beta=\alpha_0 +1$.
Clearly $|\alpha_0|<\lambda^+$ since $|\lambda\alpha_0|≤|\alpha|=\lambda$.
Now $\alpha-\alpha_0$ has cardinality smaller than $\lambda$ (else $\lambda\alpha_0+\lambda=\lambda\cdot\beta≤\alpha$), so we are done by induction on the cardinality.


Lemma 2: Let $\lambda$ some infinite cardinal, $0<\alpha<\lambda^+$ some ordinal and $X$ some set of cardinality $\lambda$. Then there is a bijection $t=(t_1,t_2)\colon\lambda\alpha\to\lambda\alpha\times X$ such that $t_1≤id_{\lambda\alpha}$.

Proof: Induction on $\alpha$. For $\alpha=1$ identify $X\sim\lambda$ and take the Mostowski collapse of the Cantor wellordering on $\lambda \times\lambda$ (given by ordering pairs first by their maximum, then by the first and then by the second component). It is known that this is a bijection $t\colon\lambda\times\lambda\leftrightarrow\lambda$ and $t_1≤id$ is clear.
Else write each $\beta<\alpha$ as $\lambda\gamma+\epsilon$ with $\gamma$ maximal (like in proof of Lemma 1.) and map it to $(\lambda\gamma + t_1(\epsilon),t_2(\epsilon))$.


Lemma 3: Let $\lambda$ some infinite cardinal, $0<\alpha<\lambda^+$ some ordinal. For every nondecresing sequence $(\kappa_i\mid i<\lambda\alpha)$ we have $\prod\limits_{i<\lambda\alpha}\kappa_i≥\left(\prod\limits_{i<\lambda\alpha}\kappa_i\right)^{\lambda\alpha}$.

Proof: We have an injection
$A:=${$f:\lambda\alpha\times\lambda\alpha\to V\mid\forall i,j <\lambda\alpha: f(i,j)\in \kappa_i$} $\to$ {$g\colon\lambda\alpha\to V\mid\forall i<\lambda\alpha: g(i)\in\kappa$}$=:B$ given by $f\mapsto f\circ t$, where $t$ is as in Lemma 2 for $X:=\lambda\alpha$.
But $\prod\limits_{i<\lambda\alpha}\kappa_i = |B|$, $\left(\prod\limits_{i<\lambda\alpha}\kappa_i\right)^{\lambda\alpha}=|A|$.


If we have a sequence of cardinals $(\kappa_i|i<\delta)$ we can determine the product as follows:

Write $\delta=\sum\limits_{k≤ l}\lambda_k\alpha_k$ as in Lemma 1. Determine the product of the first $\delta':=\sum\limits_{k≤ l-1}\lambda_k\alpha_k<\delta$ factors by induction, if $0< l$. So all that's left to do is find out what $\prod\limits_{i<\lambda_l\alpha_l}\kappa_{\delta'+i}$ is:

As before we can assume that $\delta$ is a limit, so $\lambda_l\alpha_l$ is too. Then set $\kappa=\sup\limits_{i<\lambda_l\alpha_l}\kappa_{\delta'+i}$ and then by Lemma 3: $\kappa^{\lambda_l}=\kappa^{\lambda_l\alpha_l}≤\left(\prod\limits_{i<\lambda_l\alpha_l}\kappa_{\delta'+i}\right)^{\lambda_l\alpha_l}≤\prod\limits_{i<\lambda_l\alpha_l}\kappa_{\delta'+i}≤\kappa^{\lambda_l\alpha_l}=\kappa^{\lambda_l}$

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Nice. What you really need out of Lemma 2 is that you can always partition $\lambda\alpha$ into $\lambda$ sets of size $\lambda$, each of which is cofinal. You can do this easily by partitioning each of the $\alpha$ copies of $\lambda$ separately, and then making sure each of your sets contains part of each copy of $\lambda$. –  Eric Wofsey Feb 23 '13 at 0:47

EDIT: Thanks to Eric for pointing out a stupid mistake in my proof.

So I thought about this for a little longer, and now I actually think that Stefan's Formula is true for all limit ordinals $\delta$ and strictly increasing sequences of cardinals.


Lemma 1: Let $\kappa_1≤\kappa_2$, $\lambda_1≥\lambda_2$ be infinite cardinals such that $cof(\kappa_2)≤\lambda_2$. Then $\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}=\kappa_2^{\lambda_2}$.

Proof: Obviously we have $\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}≤\kappa_2^{\lambda_1}$. We fix $\kappa_1, \lambda_1, \lambda_2$ and show "≥" by induction on $\kappa_2$ starting with the trivial case $\kappa_2=\kappa_1$.

Let's distinguish 3 cases:
case 1: $\lambda_1≥\kappa_2$: $\kappa_2^{\lambda_1}≤\lambda_1^{\lambda_1}=2^{\lambda_1}≤\kappa_1^{\lambda_1}≤\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}$
case 2: $\lambda_1<\kappa_2$ and $\forall\xi<\kappa_2:\xi^{\lambda_1}<\kappa_2$. It is known that in this case $\kappa_2^{\lambda_1}$ is either $\kappa_2$ (if $cof(\kappa_2)>\lambda_1$) or $\gimel(\kappa_2)$ (else). In both cases it is smaller (EDIT: or equal) than $\kappa_2^{\lambda_2}$
case 3: There exists some $\xi<\kappa_2$ such that $\xi^{\lambda_1}≥\kappa_2$. Then by induction $\kappa_2^{\lambda_1}≤\left(\xi^{\lambda_1}\right)^{\lambda_1}=\xi^{\lambda_1}≤\kappa_1^{\lambda_1}\xi^{\lambda_2}≤\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}$


Lemma 2: Let $\kappa_0, \kappa_l$ and $\lambda_0, \lambda_l$ be finitely many infinite cardinals such that $\forall k≤ l: cof(\kappa_k)≤\lambda_k$. Then $\max\limits_{0≤k≤ l}\kappa_k^{\lambda_k}=\prod\limits_{k=0}^l\kappa_k^{\lambda_k}=\max\limits_{0≤k,j≤ l}\kappa_k^{\lambda_j}=\left(\max\limits_{k≤ l}\kappa_k\right)^{\max\limits_{k≤ l}\lambda_k}$.

Proof: This follows immediately from Lemma 1 by induction.


Now let $\delta$ be a limit ordinal and $(\kappa(i)\mid i<\delta)$ a nondecreasing sequence of ordinals. Then writing $\delta=\lambda_0\alpha_0+\ldots+\lambda_l\alpha_l$ and $\delta_k:=\lambda_0\alpha_0+\ldots+\lambda_k\alpha_k$ as above (with $0<\alpha_k<\lambda_k^+$) we observe that all $\lambda_k≥\omega$ (since $\delta$ limit).
Then writing $\mu_k:=\sup\limits_{i<\lambda_k\alpha_k}\kappa(\delta_k+i)$ we saw $\prod\limits_{i<\delta}\kappa(i)=\max\limits_{k≤ l}\mu_k^{\lambda_k}$.
But notice that by definition $(\kappa(\delta_k+i)\mid i<\lambda_k\alpha_k)$ is cofinal in $\mu_k$, so $cof(\mu_k)≤|\lambda_k\alpha_k|=\lambda_k$. This by Lemma 2 gives us $\max\limits_{k≤ l}\mu_k^{\lambda_k}=\left(\max\limits_{k≤ l}\kappa_k\right)^{\max\limits_{k≤ l}\lambda_k}=\mu_l^{\lambda_0}=\left(\sup\limits_{i<\delta}\kappa(i)\right)^{|\delta|}$

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Tarski conjectured that for any increasing sequence $(\kappa_i\mid i\in \tau)$ of cardinals, with limit $\kappa$, we have $\prod_i \kappa_i=\kappa^{|\tau|}$. This is certainly true if $\tau$ is countable, or if $\tau$ has $|\tau|$ disjoint cofinal subsets. However, under appropriate assumptions, this fails in general for $\tau=\omega_1+\omega$, and if it fails for some $\tau$, then there is a sequence witnessing it fails for $\omega_1+\omega$. This was shown by Jech and Shelah, in "On a conjecture of Tarski on products of cardinals", PAMS 112 (4), (1991), 1117-1124. –  Andres Caicedo Feb 23 '13 at 4:32
    
Any failure (meaning, the product is strictly smaller than $\kappa^{|\tau|}$) requires a strong failure of the singular cardinal hypothesis. –  Andres Caicedo Feb 23 '13 at 4:33
    
I'm not sure if I understand your comment. surely $\prod_i \kappa_i=\kappa^{|\tau|}$ can fail if one allows repetitions in the $\kappa$'s (e.g. $\kappa_i=\aleph_i$ for $i<\omega_1$ and $\kappa_i=\aleph_{\alpha+\omega_1}$ for $\omega_1≤i<\omega$ with $\alpha >\aleph_{\omega_1}^{\aleph_1}$ under GCH) or successor $\tau$'s (e.g. my example in the comments to the first post) But doesn't my proof show that the conjecture is indeed true for strictly increasing sequences and limit $\tau$'s –  Toink Feb 23 '13 at 5:05
    
In case 2 of Lemma 1, why is $\gimel(\kappa_2)<\kappa_2^{\lambda_2}$? Indeed, if $\lambda_2<cf(\kappa_2)$, your argument shows that $\kappa_2^{\lambda_2}=\kappa_2$. –  Eric Wofsey Feb 23 '13 at 5:25
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Ah, I missed that. In that case, how can you apply the induction hypothesis to $\xi$ in case 3? You don't know that $cf(\xi)\leq\lambda_2$. –  Eric Wofsey Feb 23 '13 at 6:54

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