Jun 17, 2024 · The first induction makes sense since if he was really doing induction on $\kappa \otimes \kappa = \kappa$ for infinite cardinals, the base case would be $\omega$ (and we could disregard …

Oct 30, 2022 · So what the condition $\kappa^ {<\kappa} = \kappa$ tells us about the "size" of $\kappa$ is entirely dependent on the whims of the continuum function. At one extreme, if GCH holds, then …

1 By theorem, $\kappa (G) \leq \kappa′ (G)$ always. I need to show that $\kappa′ (G) \leq \kappa (G)$. I know that $\kappa (G)$ is the minimum size of a vertex cut. So I need to find an edge-cut that is the …

Nov 16, 2021 · 5 Not necessarily. We always have $\kappa^ {\operatorname {cf}\kappa}\le 2^\kappa,$ and, as you note, equality holds when $\kappa$ is regular (and it holds always for GCH). However, it …

May 1, 2021 · The proof relies only on properties $\kappa < \lambda$ implies $2^\kappa \leq 2^\lambda$, $\kappa < \lambda$ implies $\kappa^\mu \leq \lambda^\mu$, and connections between …

I have a few questions about Godel's pairing function and proving that $\kappa=\kappa\cdot\kappa$ for aleph cardinals. Mostly, though, I'm concerned that most of the proofs I've seen are erroneous...

Apr 21, 2012 · Under what assumptions on an infinite cardinal $\\kappa$ we have $$\\kappa^\\kappa= 2^\\kappa?$$ Please delete this question. I know the answer.

First, let me note that my definition of "strongly inaccessible" is that a nonzero ordinal $\kappa$ is strongly inaccessible if $ {\rm cf} (\kappa)=\kappa$ and $\forall\alpha<\kappa, {\cal P} …

May 25, 2017 · Without using the AC, show that if $\kappa \geq \aleph_0$ then $2^ {\kappa}-\kappa=2^ {\kappa}$. That is to say, for such $\kappa$ there is a unique cardinal $\mu$ such that …

Oct 23, 2025 · I have a trouble with a statement about cardinal exponentiation. It was stated in an answer to the Question about $\kappa^ {<\kappa} = \kappa$ Since $\kappa$ is regular, every …