# Proving equality of cardinals using cofinality.

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...

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.

• 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? Commented Jan 28 at 9:36
• @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. Commented Jan 28 at 10:41
• 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. Commented Feb 9 at 17:24

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 $$\kappa\le|\kappa^\rho|\le\kappa\implies \kappa=|\kappa^\rho|$$.