On a cardinal inequality in ZF $\mathsf {ZF}$

Pub Date : 2023-08-04 DOI:10.1002/malq.202300014

Guozhen Shen

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引用次数: 1

Abstract

It is proved in $ZF$ (without the axiom of choice) that $a^{n} ⩽ S_{n + 1} (a)$ for all infinite cardinals $a$ and all natural numbers $n \neq 0$ , where $S_{n + 1} (a)$ is the cardinality of the set of permutations with exactly $n + 1$ non-fixed points of a set which is of cardinality $a$ .

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关于ZF $\mathsf {ZF}$中的基数不等式

这是proved in 安迪是 $\ mathsf{安迪是 }$ ( 那没有选择公理》) a n ⩽ S n + 1 ( a) $\ mathfrak {a} ^ n的leqslant \ mathcal {S} {n + 1} (\ mathfrak {a })$ 为所有无限红雀队 a $\ mathfrak {a }$ 和所有自然的数字 n ≠ 0 $ n \ ne 0 $ ,哪里是n+1 (a1美元n+1美元非固定点a $ mathfrak

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