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Here are basically two questions. The first, what is the cardinal of equivalent Cauchy sequences of rationals? I know it's $\beth_1$ because of the set is essentially real numbers. But I want to know how to obtain this in view of Cauchy sequences, instead of decimal expansions. I mean, how to construct a one-to-one mapping from equivalent sequences to subsets of integers? Furthermore, what's the cardinal of Cauchy sequences? Is that ${\beth_1}^{\beth_0}=\beth_1$?

Edit

Thanks all commentators for indicating the error sourced from absence of (generalized) continuum hypothesis. Now my problem becomes asking for an explicit mapping from set of equivalent Cauchy sequences to power set of integers without computation of cardinals.

zhangwfjh
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  • Actually, it's $\beth_1$, not $\aleph_1$. $\beth_1 = 2^{\aleph_0}$, and this only equals $\aleph_1$ if the "continuum hypothesis" holds, which is something independent of the usual (ZFC) axioms of set theory. – The_Sympathizer Feb 15 '14 at 04:31
  • @mike4ty4 Did I see somewhere that "continuum hypothesis" cannot be proved nor be disproved? Then what are $\beth_0$ and $\aleph_1$? – zhangwfjh Feb 15 '14 at 04:36
  • The cardinality of the set of all Cauchy sequences is the cardinality $c$ of the continuum. The cardinality of any equivalence class is also the cardinality of the continuum. – André Nicolas Feb 15 '14 at 04:38
  • @AndréNicolas Thanks, but could you explain in detail – zhangwfjh Feb 15 '14 at 04:41
  • $\beth_0$ is $\aleph_0$, the size of the natural numbers. $\beth_1$ is $\mathfrak c$, the size of power set of the natural numbers, and the size of the reals. $\aleph_1$ is (the size of) $\omega_1$, the first uncountable ordinal, and the collection of countable ordinals. – Andrés E. Caicedo Feb 15 '14 at 04:42
  • In addition, you do not prove $\aleph_1^{\aleph_0}=\aleph_1$ in ZFC (if ZFC is consistent.) Continuum hyphothesis implies $\aleph_1^{\aleph_0}=\aleph_1$ in ZFC. However, it is consistent with ZFC that $\aleph_1^{\aleph_0}=\aleph_2$. – Hanul Jeon Feb 15 '14 at 04:48
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    Take the Cauchy sequence $1,1/2,1/4,\dots$ converging to $0$. We can modify it by choosing $a_i$ arbitrarily in the interval $(1/2^i,1/2^{i+1})$. There are at least countably many choices, even if we confine ourselves to rationals. Thus there are at least $\omega^\omega$ Cauchy sequences of rationals that converge to $0$. And since there are $c^\omega=c$ sequences, the results follow. – André Nicolas Feb 15 '14 at 04:50
  • @tetori I'm not so familiar with these, say ${\aleph_1}^{\aleph_0}=\aleph_2$. Could you recommend some books? – zhangwfjh Feb 15 '14 at 05:17
  • @AndresCaicedo So you mean there's no difference between $\aleph_0$ and $\beth_0$? Actually I didn't find $\beth$ in my logic lesson. – zhangwfjh Feb 15 '14 at 05:18
  • @AndréNicolas Thanks, that's my deduction for any Cauchy sequences. But that's your for equivalent Cauchy sequences. So these two cardinals consistent? – zhangwfjh Feb 15 '14 at 05:22
  • Do you mean that you were taught that the cardinality of the real numbers is $\aleph_1$? – Asaf Karagila Feb 15 '14 at 05:51
  • possible duplicate of What's the cardinality of all sequences with coefficients in an infinite set? – Asaf Karagila Feb 15 '14 at 06:06
  • @AsafKaragila I was taught that is $2^{\aleph_0}=\aleph_1$ – zhangwfjh Feb 15 '14 at 06:37
  • Then you were taught wrong. See this thread and also this thread. – Asaf Karagila Feb 15 '14 at 06:39
  • @AsafKaragila Thank you very much. I got that. Is there any recommendation for relative books? – zhangwfjh Feb 15 '14 at 06:43

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The set of all sequences rational numbers is $\mathbb{Q}^\mathbb{N}$, and it's cardinality is $$|\mathbb{Q}^\mathbb{N}|=|\mathbb{N}^\mathbb{N}|\leq|\mathscr{P}(\mathbb{N})^\mathbb{N}|=|\mathscr{P}(\mathbb{N}\times\mathbb{N})|=|\mathscr{P}(\mathbb{N})|,$$ where $\mathscr{P}(\mathbb{N})$ is the power set of $\mathbb{N}$. In other words:

Take two injections $\phi:\mathbb{Q}\rightarrow\mathbb{N}$ and $\psi:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ (you can construct these explicitly, using the well-ordering of $\mathbb{N}$). Then the function $\varphi:\mathbb{Q}^\mathbb{N}\rightarrow\mathscr{P}(\mathbb{N})$ given by $$\varphi(f)=\psi\left(\left\{(n,\phi(f(n))):n\in\mathbb{N}\right\}\right)$$ is injective, hence so is its restricition to the Cauchy sequences.

Luiz Cordeiro
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