# $\{ \lambda \in Card | \lambda^k > \lambda\}$ is a proper class

Suppose $k$ is an infinite cardinal. I want to prove that the classes $$A = \{\lambda \in Card | \lambda^k = \lambda\}$$ and $$B = \{\lambda \in Card | \lambda^k > \lambda\}$$ are proper. For the first class I thought that I can use fixed points of the function $F : Ord \to Ord$ where $F(\alpha) = |\alpha|^k$; since $F$ is continuous and increasing $$\forall \alpha \exists \beta>\alpha(F(\beta)=\beta)$$ Suppose A is a set (of cardinals), then $\bigcup A = \gamma$ is a cardinal and $\exists \beta>\gamma(F(\beta)=\beta)$. It's a contradiction because, since $\beta = |\beta|^k$, $\beta$ is a cardinal that belongs to $A$ and it's bigger than $\bigcup A$. Is it a correct proof? For the other class I have no idea. Some hints?

• Your $F$ is not continuous. Feb 16 '18 at 12:36
• I found another idea : for each $\lambda$ exists $\mu > \lambda$ such that $\mu^k = \mu$; namely $\mu = \lambda^k$ so $\mu \geq \lambda$ and $\mu^k = (\lambda^k)^k = \lambda^{k k} = \lambda^k = \mu$. Suppose that $A$ is a set then $\lambda = \bigcup A$ and exist, for this $\lambda$ a $\mu$ like above. Contradiction.
– M.B.
Feb 16 '18 at 15:10

You can combine two facts that you might know by now:

1. For every infinite cardinal $\mu$, $\mu^{\operatorname{cf}(\mu)}>\mu$.
2. For every regular cardinal $\kappa$, there is a proper class of cardinals with cofinality $\kappa$.

Specifically, this means that if $\operatorname{cf}(\lambda)\leq\kappa$, then $\lambda^\kappa\geq\lambda^{\operatorname{cf}(\lambda)}>\lambda$.

• The first fact is clear to me. About the second fact, I know that, for a regular cardinal $k$, the class of all ordinals with cofinality $k$ is a stationary set in $k$. First of all $k$ is not supposed to be regular in this case, am i right? Also, if there is a proper class of cardinals with cofinality $k$, is there also a proper class of cardinals with cofinality less than $k$?
– M.B.
Feb 16 '18 at 10:02
• Your second sentence makes no sense. Maybe you meant $\kappa^+$ at the end of the sentence? In any case, try to prove that fact. And obviously, if this is true for every regular $\kappa$, then it also holds for $\omega$. And while there is no proper class of cardinals with cofinality less than $\omega$, it shouldn't matter for your proof. Feb 16 '18 at 10:28
• Maybe this could work : for each infinite cardinal $\lambda$ exist $\mu>\lambda$ such that $\mu^k > \mu$. Fix a $\lambda$ infinite, then $\lambda = \aleph_\alpha$ for an $\alpha \in Ord$. I define $\mu = \aleph_{\alpha + k}$ (where the "+" is ordinal addition). $\mu^k = \aleph_{\alpha + k}^k \geq \aleph_{\alpha + k}^{cof(k)} = \aleph_{\alpha + k}^{cof(\alpha+k)} > \aleph_{\alpha + k} = \mu$ where the last inequality came from Konig theorem. Is it correct?
– M.B.
Feb 16 '18 at 15:22
• Well, the gist is fine. You should probably omit the $\aleph_{\alpha+\kappa}^{\operatorname{cof}(\alpha+\kappa)}$ part. It's pretty useless. König's theorem tells us that $\kappa^{\operatorname{cof}(\kappa)}>\kappa$, not that $\aleph_\alpha^{\operatorname{cof}(\alpha)}>\aleph_\alpha$. Feb 16 '18 at 15:28
• Ok, thank you so much
– M.B.
Feb 16 '18 at 15:41