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I'm trying to study for my Introduction to Set Theory finals, I can't seem to find an answer to this problem:

$\DeclareMathOperator{\cf}{cf}$ Suppose $\kappa, \rho$ are infinite cardinals with $\rho < \cf(\kappa)$. Suppose as well that for every cardinal $\mu < \kappa$ one has $\mu^\rho \leq \kappa$. Prove that $\kappa^\rho = \kappa$.

I'm not even sure how to approach this, if it were ordinal arithmetic it'd be rather trivial as we could just take the supremum of a set $\leq \kappa$, but I am much less used to cardinal arithmetic (specially when cofinalities are involved). Any good resources on how I could learn how to solve problems such as this one would also be appreciated...

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2 Answers 2

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For any function $f: \rho \to \kappa$ we have that its image must be contained in some $\mu < \kappa$, because $\rho < \operatorname{cf}(\kappa)$. Hence $$ \kappa^\rho = \bigcup_{\mu < \kappa} \mu^\rho. $$ So in terms of cardinal arithmetic we then have that $\kappa^\rho = \sup_{\mu < \kappa} \mu^\rho$. As $\mu^\rho \leq \kappa$ for all $\mu < \kappa$ we thus have that $\kappa^\rho = \sup_{\mu < \kappa} \mu^\rho \leq \kappa$. Clearly $\kappa \leq \kappa^\rho$ also holds, and the result follows.

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  • $\begingroup$ Hey, I can't seem to figure out how you deduced that $\kappa^\rho = \bigcup_{\mu < \kappa} \mu^\rho$, which function $f$ are you considering? $\endgroup$
    – TC159
    Commented Jan 28 at 9:36
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    $\begingroup$ @TC159 I'm not considering a single function $f$. In the first sentence I argue that for all $f \in \kappa^\rho$ there is $\mu < \kappa$ such that $f \in \mu^\rho$. Hence $\kappa^\rho \subseteq \bigcup_{\mu < \kappa} \mu^\rho$. The other inclusion always holds. $\endgroup$ Commented Jan 28 at 10:41
  • $\begingroup$ Great news, I read this carefully a while back and think that made me finally understand how to use cofinalities. I had my final exam today and it went extremely well, thank you. $\endgroup$
    – TC159
    Commented Feb 9 at 17:24
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Here are some references for this from Thomas Jech 'Set Theory' 3rd edition

Lemma $3.3$ part $(3.9)$ If $0<\lambda\le\mu$ then $|\kappa^\lambda|\le|\kappa^\mu|$. Since $0<1\le\rho$, $\kappa=|\kappa^1|\le|\kappa^\rho|$.

Lemma $3.9(\text{ii})$ If $\lambda<\text{cf}(\kappa)$ and $f\colon\lambda\to\kappa$ then $\bigcup\text{range}(f)<\kappa$. Since $\rho<\text{cf}(\kappa)$ then $f\in\kappa^\rho$ has $\bigcup\text{range}(f)<\kappa$ so taking $\mu=\bigcup\text{range}(f)+1<\kappa$ we have $\text{range}(f)\subseteq\mu$.

If $\mu<\kappa$ and $f\in\mu^\rho$ then $f\subseteq\rho\times\mu\subseteq\rho\times\kappa$ so $f\in\mathcal{P}(\rho\times\mu)\subseteq\mathcal{P}(\rho\times\kappa)$ and we can view $\mu^\rho$ as a subset of $\kappa^\rho$. From the previous paragraph it follows that $\bigcup_{\mu<\kappa}\mu^\rho=\kappa^\rho$.

Now Lemma $5.2$ $|\bigcup S|\le|S|\cdot\sup\{|X|\colon X\in S\}$. This lemma assumes a previous result Lemma $3.4(\text{ii})$ that if $X$ is a set of cardinals then $\sup X=\bigcup X$ is a cardinal. Unlike previous statements, Lemma $5.2$ depends on the axiom of choice.

Applying this to $\bigcup_{\mu<\kappa}\mu^\rho=\kappa^\rho$ $$|\kappa^\rho|\le |\kappa|\cdot\sup\{|\mu^\rho|\colon\mu<\kappa\}\tag{1}$$

Since we're given that $|\mu^\rho|\le\kappa$ for all $\mu<\kappa$ and the $\sup$ as ordinals of a set of cardinals is a cardinal it follows that $\sup\{|\mu^\rho|\colon\mu<\kappa\}\le\kappa$. Also $|\kappa|=\kappa$ so from $(1)$ we get $$|\kappa^\rho|\le\kappa\cdot\kappa=|\kappa\times\kappa|=\kappa$$ where $\kappa\cdot\kappa=\kappa$ is Theorem $3.5$.

From the Dedekind-Bernstein theorem $3.2$ we have $\kappa\le|\kappa^\rho|\le\kappa\implies \kappa=|\kappa^\rho|$.

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