1
$\begingroup$

Let $\{x_n\}$ be a sequence. Suppose that there are two convergent subsequences $\{x_{n_i}\}$ and $\{x_{m_i}\}$. Suppose that $\lim_{i\to\infty} x_{n_i} =a$ and $\lim_{i\to\infty} x_{m_i} =b$ where $a \neq b$. Prove that $\{x_n\}$ is not convergent.

I cannot use the fact that if a sequence is convergent, then its subsequences are convergent and the limit of the sequence is equal to the limit of the subsequence. I am stuck figuring out how to prove that it is not convergent a different way.

$\endgroup$
1
  • 1
    $\begingroup$ hint - use Cauchy criterion. for large $i$ both $x_{n_i}$ and $x_{m_i}$ are close to their respective limits, so cannot be close to eachother $\endgroup$
    – mm-aops
    Jun 26, 2013 at 13:34

1 Answer 1

Reset to default
2
$\begingroup$

Well, in a way any proof will use this fact. Anyways, the hint is to assume that $\lim x_n = c$ does exist. Then at least $c\neq a$ or $c\neq b$ or both, so without loss of generality we can assume that $c\neq a$. Let $\varepsilon = \frac13|c-a|$, and find $N_1$ such that $|x_n-c|<\varepsilon$ for $n>N_1$, and $N_2$ such that $|x_{n_i}-a|<\varepsilon$ for $i>N_2$. I hope, you can see a contradiction here now.

$\endgroup$
4
  • $\begingroup$ what do I do to find N1 and N2? $\endgroup$
    – user72195
    Jun 30, 2013 at 21:35
  • $\begingroup$ @user72195 from th definition of converging subsequences/sequences $\endgroup$
    – Ilya
    Jul 1, 2013 at 7:50
  • $\begingroup$ @Ilya I am below 75 IQ. What is the contradiction? $\endgroup$
    – MeditationOrBust
    Feb 2 at 20:01
  • $\begingroup$ @meditationorbust just pick epsilon smaller than half difference between a and c $\endgroup$
    – Ilya
    Feb 4 at 0:42

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.