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After having proven that:

  • every Cauchy sequence is bounded
  • that for any positive real numbers $x$ and $y$ there is some natural number $k\in {\mathbb N}$ such that $kx > y$
  • that for any positive real number $x$ there is some rational number $b\in {\mathbb Q}$ such that $0 < b < x$
  • and the theorem that every Cauchy sequence of rational numbers $\{a_n\}_{n\in {\mathbb N}}$ is convergent in $\mathbb R$ with the limit $\lim_{n \rightarrow \infty} a_n = [(a_n)]$ where $[a_n]$ has been defined as the equivalence class of equivalent Cauchy sequences in the set of rational numbers i.e. $[(a_n)] = \{{b_n} \in A│{{b_n}} \sim {{a_n}}\}$, where $A$ is the set of all Cauchy sequences in $\mathbb Q$ and $a_n \in A$,

I am trying to next prove as a corollary that for a real number $[(a_n)]$ and a positive real number $\epsilon$ , there is a rational number $a$ such that $|[(a_n)] - a|<\epsilon$. Equivalent sequences are defined by $[{{a_n}}\sim {{b_n}}] \equiv [\lim(a_n - b_n) = 0]$ How do I prove this corollary from the previous propositions given?

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  • $\begingroup$ Your question is incomplete? $\endgroup$ Commented May 19, 2017 at 2:01
  • $\begingroup$ Hi Eric, please format your question into one coherent piece instead of splitting it up in the comments to your question for your readers' sakes. $\endgroup$ Commented May 19, 2017 at 3:18
  • $\begingroup$ I formatted the question into one piece. $\endgroup$ Commented May 19, 2017 at 4:29
  • $\begingroup$ I guess you are trying to base your question on the assumption that real numbers are constructed as equivalence classes of Cauchy sequences of rationals. $\endgroup$ Commented May 19, 2017 at 8:22
  • $\begingroup$ Yes, that is correct. $\endgroup$ Commented May 19, 2017 at 8:31

3 Answers 3

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Suppose that [{an}] is any real number and ϵ is any positive real number. By definition of [{an}], {an} is a Cauchy sequence in Q. Therefore, by the theorem that every Cauchy sequence of rational numbers {an}is convergent in R with the limit lim an=[(an)]. Now by definition of the limit, |an -[{an}]|<ϵ for every n >N and some N∈N. Consequently, |[{an}]-an|<ϵ for every n >N and some N∈N. Therefore, setting an = a for every n >N and some N∈N, there is a rational number a such that |[{an}]-a|<ϵ.

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The idea is simple. Since $\epsilon$ is a positive real number it can be represented as $\epsilon=[e_{n}] $ where $e_{n} $ is an increasing sequence of positive rational numbers. Next we choose integers $M_{k} >0$ such that $|a_{m} - a_{n} |<e_{k} /2$ for $k=1,2,\dots$ and $m, n$ both greater than or equal to $M_{k} $. It is sufficient to choose the rational number $a=a_{M_{k}} $ for any fixed $k$ and it will satisfy $|[a_{n}] - a|\leq e_{k} /2<\epsilon$.

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First prove that no positive real is less than every positive rational.

So when $0\leq a<b,$ take $q_0\in \mathbb Q$ with $0<q_0<\frac {b-a}{3}.$ Let $k_0$ be the least $k\in \mathbb N$ such that $kq_0>a.$ So $(k_0-1)q_0\leq a<k_0q_0.$ Therefore $$a<(k_0+1)q_0=(k_0-1)q_0+2q_0\leq a+2q_0<a+2\left(\frac {b-a}{3}\right)<b.$$

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  • $\begingroup$ This does not prove that for a real number [(an)] and a positive real number ϵ , there is a rational number a such that |[(an)]−a|<ϵ. $\endgroup$ Commented May 20, 2017 at 0:45

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