A Choice-Free Cardinal Equality

IF 0.6 3区 数学 Q2 LOGIC
G. Shen
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引用次数: 0

Abstract

For a cardinal $\mathfrak{a}$, let $\mathrm{fin}(\mathfrak{a})$ be the cardinality of the set of all finite subsets of a set which is of cardinality $\mathfrak{a}$. It is proved without the aid of the axiom of choice that for all infinite cardinals $\mathfrak{a}$ and all natural numbers $n$, \[ 2^{\mathrm{fin}(\mathfrak{a})^n}=2^{[\mathrm{fin}(\mathfrak{a})]^n}. \] On the other hand, it is proved that the following statement is consistent with $\mathsf{ZF}$: there exists an infinite cardinal $\mathfrak{a}$ such that \[ 2^{\mathrm{fin}(\mathfrak{a})}<2^{\mathrm{fin}(\mathfrak{a})^2}<2^{\mathrm{fin}(\mathfrak{a})^3}<\dots<2^{\mathrm{fin}(\mathrm{fin}(\mathfrak{a}))}. \]
自由选择的基本平等
对于基数$\mathfrak{a}$,设$\mathrm{fin}(\mathfrak{a})$为基数为$\mathfrak{a}$的集合的所有有限子集的集合的基数。不借助于选择公理,证明了对于所有无限基数$\mathfrak{a}$和所有自然数$n$, \[ 2^{\mathrm{fin}(\mathfrak{a})^n}=2^{[\mathrm{fin}(\mathfrak{a})]^n}. \],另一方面,证明了下列命题与$\mathsf{ZF}$一致:存在一个无限基数$\mathfrak{a}$,使得 \[ 2^{\mathrm{fin}(\mathfrak{a})}<2^{\mathrm{fin}(\mathfrak{a})^2}<2^{\mathrm{fin}(\mathfrak{a})^3}<\dots<2^{\mathrm{fin}(\mathrm{fin}(\mathfrak{a}))}. \]
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来源期刊
CiteScore
1.00
自引率
14.30%
发文量
14
审稿时长
>12 weeks
期刊介绍: The Notre Dame Journal of Formal Logic, founded in 1960, aims to publish high quality and original research papers in philosophical logic, mathematical logic, and related areas, including papers of compelling historical interest. The Journal is also willing to selectively publish expository articles on important current topics of interest as well as book reviews.
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