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Suppose $\kappa$ and $\lambda$ are infinite cardinals and that $\lambda$ is regular. Kunen states somewhere that this means that we have $$\left(\kappa^{<\lambda}\right)^{<\lambda}=\kappa^{<\lambda}$$ I want to prove this. The inequality $\geq$ is clear, so let's focus on the other one. We can also assume $\lambda$ is a limit cardinal. I'm going to mix arithmetic and functional notation a bit, because I find it a bit easier to think this way. So, we're looking at $$\left(\kappa^{<\lambda}\right)^{<\lambda}=\bigcup_{\mu<\lambda}\left(\mu\to\left(\bigcup_{\nu<\lambda}\left(\nu\to\kappa\right)\right)\right)=(*)$$ Since $\lambda$ is regular, for a fixed $\mu$ in the first union, the second union can be shortened to a $\bigcup_{\nu<\zeta(\mu)}$, for some cardinal $\mu\leq\zeta(\mu)<\lambda$. We can now continue the above equality with $$(*)=\bigcup_{\mu<\lambda}\left(\mu\to\sum_{\nu<\zeta(\mu)}\kappa^\nu\right)\leq \bigcup_{\mu<\lambda}\left(\mu\to\zeta(\mu)\cdot\kappa^{\zeta(\mu)}\right)=\sum_{\mu<\lambda}\zeta(\mu)^\mu\cdot\kappa^{\zeta(\mu)}$$

I don't know how to continue from this point. The expression on the right looks a bit like what I want to get, but I feel like my estimates were a bit too rough. I'm also worried that $\zeta(\mu)$ is unbounded in $\lambda$, but I'm sure something of this sort must come into play, because I don't see any other way to use the fact that $\lambda$ is regular.

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3 Answers 3

up vote 1 down vote accepted

Suppose that $\sigma\in^{<\lambda}\left(^{<\lambda}\kappa\right)$, say $\sigma:\mu\to^{<\lambda}\kappa$. Let $\alpha=\sup\{\operatorname{dom}\,\sigma_\xi:\xi<\mu\}<\lambda$. For $\xi<\mu$ define $$\hat\sigma_\xi:\alpha\to\kappa:\zeta\mapsto\begin{cases} \sigma_\xi(\zeta)+1,&\text{if }\zeta\in\operatorname{dom}\,\sigma_\xi\\ 0,&\text{otherwise}\;. \end{cases}$$

Note that $\sigma_\xi$ can be recovered from $\hat\sigma_\xi$. Define

$$\bar\sigma:\mu\times\alpha\to\kappa:\langle\xi,\zeta\rangle\mapsto\hat\sigma_\xi(\zeta)\;;$$

clearly $\sigma$ can be recovered from $\bar\sigma$. Thus, it suffices to show that $$\left|\bigcup_{\mu,\alpha<\lambda} {^{(\mu\times\alpha)}\kappa}\right|\le\left|\bigcup_{\mu<\lambda}{^\mu\kappa}\right|\;.$$

But

$$\begin{align*} \left|\bigcup_{\mu,\alpha<\lambda} {^{(\mu\times\alpha)}\kappa}\right|&=\left|\bigcup_{\mu<\lambda}\;\bigcup_{\alpha,\beta\le\mu}{^{(\alpha\times\beta)}\kappa}\right|\\ &\le\left|\bigcup_{\mu<\lambda}\;\bigcup_{\alpha\le\mu}{^\mu\kappa}\right|\\ &=\left|\bigcup_{\mu<\lambda}\left(\mu\times(^\mu\kappa)\right)\right|\\ &=\left|\bigcup_{\mu<\lambda}{^\mu\kappa}\right|\;. \end{align*}$$

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Thanks. Your and David's answers are similar; I accepted yours because it lines up a bit more with what I was (unsuccessfully) trying to do. –  Miha Habič Jan 27 '12 at 12:01

I will be using Introduction to Cardinal Arithmetic. (Holtz, Steffens, Weitz. 1999, Birkhäuser). I'll cite (up to minor alterations) two claims and parts of the proofs. I will skip the parts that seem obvious to me, but feel free to ask for details. $\newcommand{\cf}{\operatorname{cf}}$

Lemma 1.6.15(c): If $\lambda$ and $\kappa$ are cardinal numbers such that $\kappa\ge\aleph_0$ and $0<\lambda<\cf(\kappa)$, then $$\kappa^\lambda=\kappa\cdot\sup\{\nu^\lambda:v<\kappa\}=\kappa\cdot\sum_{\nu<\kappa}\nu^\lambda$$

The proof is relatively straightforward, so I will not include it in here. We use this to prove the following:

Lemma 1.7.6(c): Assume that $\kappa$ and $\lambda$ are cardinal numbers such that $\kappa\ge 2$ and $\lambda\ge\omega$.

If $\nu>0$ is a cardinal number, then $$(\kappa^{<\lambda})^\nu=\begin{cases}\kappa^{<\lambda} & if\quad 0<\nu<\cf(\lambda),\\ \kappa^\lambda & if\quad \cf(\lambda)\ge\nu<\lambda,\\ \kappa^\nu & if\quad \lambda\le\nu.\end{cases}$$

In our case we have:

$$\left(\kappa^{<\lambda}\right)^{<\lambda}=\sum_{\mu<\lambda}\left(\kappa^{<\lambda}\right)^\mu\stackrel{*}{=}\sum_{\mu<\lambda}\left(\kappa^{<\lambda}\right)=\sum_{\mu,\nu<\lambda}\kappa^{\mu\cdot\nu}=\left(\kappa^{<\lambda}\right)$$

Where $\stackrel{\ast}{=}$ is exactly the first case, since $\lambda$ is regular and $\cf(\lambda)=\lambda$. It is indeed enough to prove the first case:

We have two cases, if the continuum function for $\kappa$ ($\mu\mapsto\kappa^\mu$) is eventually constant below $\lambda$ then for some $\rho>\mu$ we have: $$\left(\kappa^{<\lambda}\right)^\mu = \left(\kappa^\rho\right)^\mu = \kappa^\rho=\kappa^{<\lambda}$$

If the continuum function for $\kappa$ is not eventually constant below $\lambda$ then $\kappa^{<\lambda}=\sup\{\kappa^{\mu_\xi}\mid \mu,\xi<\lambda\}$ where both $\mu_\xi$ and $\kappa^{\mu_\xi}$ are strictly increasing sequences. Without loss of generality we can assume that for all $\xi$, $\mu<\mu_\xi$.

Since we have a strictly increasing sequence of length $\lambda$ we have that $\cf(\kappa^{<\lambda})=\cf(\lambda)=\lambda>\mu$ we have by a Lemma 1.6.15(c):

$$\left(\kappa^{<\lambda}\right)^\mu=\kappa^{<\lambda}\cdot\sum_{\xi<\lambda}(\kappa^{\mu_\xi})^\mu=\sum_{\xi<\lambda}(\kappa^{\mu_\xi})=\kappa^{<\lambda}$$

Note that in our case $\kappa^{<\lambda}$ is the $\kappa$ of the previous lemma, and thus indeed the conditions hold we can also replace the sum in the previous lemma by any other cofinal sum, as you should know by now.

So we concludes our proof for the evening.

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Great answer Asaf. I accepted Brian's because it's a bit more conceptual, but I'll want to internalize these methods so that I can carry out arithmetic a bit more like you show. And thanks for the reference. –  Miha Habič Jan 27 '12 at 12:04
    
@Miha: It's fine. The approach I used here is not common and I only resorted to it due to lack of success in the approach Brian used. It is interesting to know these rules, though. I don't know them and often use the book as reference. –  Asaf Karagila Jan 27 '12 at 14:13

An element $f$ of $^{<\lambda}\left(^{<\lambda}\kappa\right)$ can be viewed as a map from a subset $\{(\alpha,\beta)\mid \alpha<\mu, \beta<\nu_{\alpha}\}$ of $\lambda\times\lambda$ to $\kappa$, where $\mu<\lambda$ and, for $\alpha<\mu$, $\nu_{\alpha}<\lambda$. By regularity, such a subset is contained in $\mu\times\mu'$ for some $\mu'<\lambda$. Now, we can find a bijection $\phi$ from $\lambda$ onto $\lambda\times\lambda$ so that any $\mu\times\mu'$ is contained in the image of a proper initial segment $\gamma$ of $\lambda$ (for example, see Theorem I.10.12 in Kunen's book.) Fixing any $\bot\notin\kappa$, we can extend $f\circ\phi$ to an element $f'$ of $^\gamma(\kappa\cup\{\bot\})$ by filling in the otherwise undefined elements of the domain with $\bot$. Then sending $f$ to $f'$ gives an injection from $^{<\lambda}\left(^{<\lambda}\kappa\right)$ into $^{<\lambda}(\kappa\cup\{\bot\})$. Since $\kappa$ is infinite, $\kappa$ and $\kappa\cup\{\bot\}$ have the same cardinality, so $\left(\kappa^{<\lambda}\right)^{<\lambda}\le \kappa^{<\lambda}$.

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