0
$\begingroup$

Let $\langle a_n\rangle$ , $\langle b_n\rangle$ , $\langle c_n\rangle$ be Cauchy sequences of rational numbers, and $\langle c_n\rangle$ is equivalent to $\langle a_nb_n\rangle$. Prove or disprove that there are two Cauchy sequences $\langle a_n'\rangle$ , $\langle b_n'\rangle$ of rational numbers such that

(1) $\langle a_n\rangle$ is equivalent to $\langle a_n'\rangle$ ;

(2) $\langle b_n\rangle$ is equivalent to $\langle b_n'\rangle$ ;

(3) $\langle c_n\rangle=\langle a_n'b_n'\rangle$ .

If it is true, can we prove it intuitionistically?

$\endgroup$
4
  • 1
    $\begingroup$ Why don't you let $<a_n>=<a'_n>$ and $<b_n>=<b'_n>$ $\endgroup$ Commented Dec 2, 2012 at 11:32
  • $\begingroup$ @Amr <c_n> is equivalent to <a_nb_n> but may be not equal. $\endgroup$ Commented Dec 2, 2012 at 11:37
  • $\begingroup$ oh i see. I thought you were using = as equivalent $\endgroup$ Commented Dec 2, 2012 at 11:38
  • $\begingroup$ Does this not follow immediately from the transitivity property for equivalence relations? $\endgroup$ Commented Dec 2, 2012 at 12:11

1 Answer 1

Reset to default
1
$\begingroup$

Case 1: both $a_n,b_n$ converge to zero. Let $a'_n=r_n,b'_n=c_n/r_n$. Where $r_n$ is a sequence of rationals such that for all n $|r_n-\sqrt c_n|<1/n$

Case 2: one of the two sequences does not converge to zero

Let $lim_{n\rightarrow \infty} a_n=A$ . Assume WLOG that A is different from $0$. Let $a'_n$ be a sequence of nonzero rationals converging to $A$, and let $b'_n=c_n/a'_n$

$\endgroup$
5
  • $\begingroup$ If A=0 , we can't prove that <b_n> is equivalent to <b'_n>. $\endgroup$ Commented Dec 2, 2012 at 12:05
  • $\begingroup$ yes i will edit my answer $\endgroup$ Commented Dec 2, 2012 at 12:07
  • $\begingroup$ In case 1 , we can't prove that for all n , $c_n=a_n'b_n'$ . $\endgroup$ Commented Dec 2, 2012 at 12:25
  • $\begingroup$ I edited it again ! $\endgroup$ Commented Dec 2, 2012 at 15:40
  • $\begingroup$ In case 1, how can we prove that $b_n'\rightarrow 0$ ? $\endgroup$ Commented Dec 2, 2012 at 18:10

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.