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    $2^\\kappa-\\kappa=2^\\kappa$ for infinite $\\kappa$ without the AC

    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 $\kappa+\mu=2^{\kappa}$, ...

    logic - Why is $\gimel (\kappa)>\kappa$? - Mathematics Stack …

    Dec 23, 2023 · I want to prove that $\gimel(\kappa)=\kappa^{\text{cof}(\kappa)}>\kappa$, by using König's theorem. So I basically said this: $$\kappa^{\text{cof}(\kappa)}=\prod_{i ...

    Cardinal power $\\kappa^\\kappa$. When is it equal to $2^\\kappa$?

    Assuming $\kappa$ is an ordinal the answer is always. The reason is simple: by Cantor's theorem we have $2<\kappa<2^\kappa$, therefore using exponentiation laws: $$2^\kappa\le\kappa^\kappa\le\left(2^\kappa\right)^\kappa=2^{\kappa\cdot\kappa}=2^\kappa$$

    elementary set theory - Show that for any infinite cardinal $\kappa ...

    Jun 27, 2020 · $\begingroup$ I know that subtraction isn't necessarily well defined for infinite cardinals, but since $\kappa$ is strictly less than $2^{\kappa}$, then by the absorption principle, $2^{\kappa}= 2^{\kappa} + \kappa$. $\endgroup$

    set theory - Question about $\kappa^{<\kappa} = \kappa

    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 every regular cardinal satisfies $\kappa^{<\kappa} = \kappa$.

    set theory - If $ \kappa - Mathematics Stack Exchange

    If $ \kappa $ is strongly inaccessible, then $ \kappa^{<\kappa} = \kappa $ Strongly inaccessible means that $ \kappa $ is uncountable, limit regular and for all $ \lambda < \kappa $ it holds that $ 2^\lambda < \kappa $ I think that it is obvious that $ \kappa^{<\kappa} \geq \kappa $ for infinite cardinals, so I'm going to pursue the other ...

    set theory - Weakly inaccessible cardinal such that …

    Jun 16, 2019 · The simplest example I can come up with is that it is consistent relative to some very mild assumptions (I think the consistency of one inaccessible suffices) that $\mathfrak c$ is weakly inaccessible and that $2^\alpha=\mathfrak c$ for all $\alpha < \mathfrak c.$ In this case $\kappa^{<\kappa}=\kappa$ holds for $\kappa=\mathfrak c$ and clearly ...

    Why $2^\\kappa=\\kappa^{\\operatorname{cf}{\\kappa}}$, if …

    Another fact worth mentioning is: If $\kappa$ is a strong limit cardinal, $2^\kappa=\kappa^{\operatorname{cf}{\kappa}}$. My question is how to derive this "fact"? My first guess is that strong limit cardinals are regular, but it's obviously wrong, since if so, definition of inaccessible cardinal would be redundant.

    matrices - If $\kappa (A) > \kappa (B)$, show $\kappa (B^ {-1}A ...

    Jun 1, 2019 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

    solution verification - If $\kappa=\kappa^{<\kappa}$ then there is a ...

    Aug 11, 2021 · If $\kappa=\kappa^{<\kappa}$ then there exists a dense linear order of size $\kappa$ in which every ordinal of cardinality $\leq\kappa$ can be embedded.