Density of Cauchy sequences of rational numbers 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?
 A: 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.
A: 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 $|[a_{n}] - a|\leq e_{k} /2<\epsilon$.
A: 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 $$a<(k_0+1)q_0=(k_0-1)q_0+2q_0\leq a+2q_0<a+2\left(\frac {b-a}{3}\right)<b.$$
