19 Sets, functions, and the continuum hypothesis
The set of \(\mathbb {Q}\) of rational numbers is countable.
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The set \(\mathbb {R}\) of real numbers is not countable
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The set \(\mathbb {R}^2\) of all ordered pairs of real numbers (that is, the real plane) has the same size as \(\mathbb {R}\).
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If each of two sets \(M\) and \(N\) can be mapped injectively into the other, then there is a bijection from \(M\) to \(N\), that is \(|M| = |N|\).
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If \(c {\gt} \aleph _1\), then every family \(\{ f_\alpha \} \) satisfying \((P_0)\) is countable. If, on the other hand, \(c = \aleph _1\), then there exists some family \(\{ f_\alpha \} \) with property \(P_0\) which has size \(c\).
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Appendix: On cardinal and ordinal numbers
Let \(\mu \) be an ordinal number and denote by \(W_\mu \) the set of ordinal numbers smaller than \(\mu \). Then the following holds:
The elements of \(W_\mu \) are pairwise comparable.
If we order \(W_\mu \) according to their magnitude, then \(W_\mu \) is well-ordered and has ordinal number \(\mu \).
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Any two ordinal numbers \(\mu \) and \(\nu \) satisfy precisely one of the relations \(\mu {\lt} \nu \), \(\mu = \nu \), or \(\mu {\gt} \nu \).
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Every set of ordinal numbers (ordered according to magnitude) is well-ordered.
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For every cardinal number \(\mathfrak {m}\), there is a definite next larger cardinal number.
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Let the infinite set \(M\) have cardinality \(\mathfrak {m}\), and let \(M\) be well ordered according to the initial ordinal number \(\omega _{\mathfrak {m}}\). Then \(M\) has no last element.
Indeed, if \(M\) had a last element \(m\), then the segment \(M_m\) would have an ordinal number \(\mu {\lt} \omega _{\mathfrak {m}}\) with \(|\mu | = \mathfrak {m}\), contradicting the definition of \(\omega _{\mathfrak {m}}\).
Suppose \(\{ A_\alpha \} \) is a family of size \(\mathfrak {m}\) of countable sets \(A_\alpha \), where \(\mathfrak {m}\) is an infinite cardinal. Then the union \(\bigcup _\alpha A_\alpha \) has size at most \(\mathfrak {m}\).
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