1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Your question is incomplete? $\endgroup$ – Juanito May 19 '17 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. $\endgroup$ – Alex Ortiz May 19 '17 at 3:18
  • $\begingroup$ I formatted the question into one piece. $\endgroup$ – Eric Brown May 19 '17 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. $\endgroup$ – Paramanand Singh May 19 '17 at 8:22
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Eric Brown May 19 '17 at 8:31
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$a<(k_0+1)q_0=(k_0-1)q_0+2q_0\leq a+2q_0<a+2\left(\frac {b-a}{3}\right)<b.$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Eric Brown May 20 '17 at 0:45
0
$\begingroup$

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 $|[a_{n}] - a|\leq e_{k} /2<\epsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.