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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.

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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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? $\endgroup$ 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

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