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I'm not even sure how to approach this problem. Kunen defined for arbitary sets $B$ and cardinals $\theta$, $$ [B]^\theta = \{ x \subseteq B : |x| = \theta \} $$ $$ [B]^{< \theta} = \{ x \subseteq B : |x| \leq \theta \} $$ And the problem (working in ZFC without foundation) I'm stuck on is the second part of lemma I.13.19: if $B$ is infinite and $\aleph_{0} \leq \theta \leq |B|$, then $$ |[B]^\theta | = |B|^\theta $$ And $$ |[B]^{< \theta} | = |B|^{< \theta} $$ Here $B^A$ is the cardinal exponentiation, and $B^{< A} = | \{ f: C \to B : C$ is an ordinal and $C < A\}|$ I've only managed to show the trivial inequalities like $|[B]^\theta | \leq |B|^\theta$. I'm sure the proof is probably pretty simple, but I'm pretty clueless about what I should do next, any hints would be appreciated!

Mark Kamsma
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SpooFwen
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  • What was your proof for the statement that you were able to prove? – Michael McGovern Jun 13 '19 at 00:49
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    You might first try the case $\theta=|B|=\aleph_0$. – Eric Wofsey Jun 13 '19 at 00:52
  • Which book is that? I have Kunen, Set Theory: An Introduction to Independence Proofs (Elsevier, 1980), and there chapter I §13 doesn't contain anything numbered higher than 13.2. – hmakholm left over Monica Jun 13 '19 at 00:53
  • @Henning Makholm it's his 2013 set theory book, I think he has a related exercise at the end of his section on cardinal arithmetic from his foundations of mathematics book – SpooFwen Jun 13 '19 at 00:54
  • @Michael McGovern if you are talking about the inequality, I figired I could just make an injection from subsets of size $\theta$ to some bijection from that set to $\theta$, if you are talking about the first half of the lemma, it was to show that $| [B]^n| =|[B]^{< \omega}| = |B|$ I figured $| [B]^n| = |B×(B-1)×...×(B-n+1)| = |B|$ where $(B-k)$ is just B with k less elements. And $|[B]^{< \omega}|$ is just $\omega$ blocks of things of cardinality $|B|$, so its cardinality must be $\omega \cdot |B|$ – SpooFwen Jun 13 '19 at 01:06
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    I am pretty sure $[B]^{< \theta}$ should be all the subsets of $B$ of cardinality strictly less than $\theta$ (your third line says "less then or equal to"). – Mark Kamsma Jun 13 '19 at 10:47

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