Given $\{a_n\}_{n=0}^\infty$ and $\{b_n\}_{n=0}^\infty$ bounded sequences; show that if $\lim \limits_{n\to \infty}a_n-b_n=0$ then both sequences have the same subsequential limits.

My attempt to prove this begins with: Let $E_A=\{L|L$ subsequential limit of $a_n$} and $E_B=\{L|L$ subsequential limit of $b_n$}. We need to show that $E_A=E_B$.

Given bounded sequence $a_n$ and $b_n$ we know from B.W that each sequence has a subsequence that converges, therefore both $E_A$ and $E_B$ are not empty;

Let $L\in E_A$. How can I show that $L\in E_B$?

Thank you very much.

  • $\begingroup$ You can start by proving the result that $\limsup (x_{n}+y_{n})=\lim x_{n}+\limsup y_{n}$ whenever the sequence $(x_n)$ is convergent, see this question. $\endgroup$ – Martin Sleziak May 1 '12 at 16:10

You can approach it very directly. Let $L\in E_A$. Then there is a subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$ converging to $L$; now use the hypothesis that $\langle a_n-b_n:n\in\Bbb N\rangle\to 0$ to show that $\langle b_{n_k}:k\in\Bbb N\rangle$ converges to $L$.

Note that you really only have to prove that $E_A\subseteq E_B$: if $\langle a_n-b_n:n\in\Bbb N\rangle\to 0$, then clearly also $\langle b_n-a_n:n\in\Bbb N\rangle\to 0$, and you can appeal to the first half to conclude that $E_B\subseteq E_A$.

  • $\begingroup$ how from $\forall \epsilon>0, \exists n_0,\forall n>n_0, |a_n-b_n-L|<\epsilon$ can I deduce that ${b}_{n_k}$ converges to L? $\endgroup$ – Anonymous May 1 '12 at 16:24
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    $\begingroup$ @Anonymous: $|b_{n_k}-L|=|(b_{n_k}-a_{n_k})+(a_{n_k}-L)|\le|b_{n_k}-a_{n_k}|+|a_{n_k}-L|$. $\endgroup$ – Brian M. Scott May 1 '12 at 16:27

I think you may want to prove that:

  1. Consider two sequences $\{x_n\}$ and $\{y_n\}$ such that $x_n-y_n \to l$ and $y_n \to y$, then, $$x_n \to l+y$$

  2. Given a convergent sequence $\{x_n\}$ that converges to $x$, all its subsequences converge to the same limit, $x$.

Do you see how that would pay here?

  • Let $r \in E_B$. That is, $r$ is a limit point of the sequence $\{b_n\}$. So, there is a subsequence of $\{b_n\}$, say $\{b_{n_k}\}$ that converges to $r$.

  • Now, consider the same subsequence of $\{a_n-b_n\}$, namely $\{a_{n_k}-b_{n_k}\}$. Since this is a subsequence of a convergent subsequence, $\{a_{n_k}-b_{n_k}\}$ converges to $0$, $a_{n_k}-b_{n_k} \to 0$ by $(2)$.

Now putting together the two claims, by $(1)$. you have that $a_{n_k} \to r$. That is, $r \in E_A$. This proves one inclusion, $E_B \subseteq E_A$. The proof of the other inclusion is similar.

For the other inclusion, as Brian observes, note that $a_n -b_n \to 0$ implies $b_n-a_n \to 0$. Now appeal to the previous part, to see that $E_A \subseteq E_B$.

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    $\begingroup$ I think that should be $x+b$, not $x-b$. $\endgroup$ – Cameron Buie May 1 '12 at 16:00
  • $\begingroup$ @CameronBuie That was an unfortunate typo, sorry! $\endgroup$ – user21436 May 1 '12 at 16:03
  • $\begingroup$ I'm familiar with the two claims; however, in the question above it isn't given that $b_n$ converges, am I wrong? $\endgroup$ – Anonymous May 1 '12 at 16:08
  • $\begingroup$ @Kannappan Sampath Thank you very much for your help, I'm waiting patiently. $\endgroup$ – Anonymous May 1 '12 at 16:16
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    $\begingroup$ @Anonymous What would a subsequence of $\{c_n\}$ look like? Now set $c_n:=a_n-b_n$, if there was a term $c_{n_k}$ in your subsequence, the corresponding term would be $c_{n_k}=a_{n_k}-b_{n_k}$, right? $\endgroup$ – user21436 May 1 '12 at 16:41

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