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Let $c$ be a limit point of a sequence of real numbers $\langle a_n \rangle$ and $d$ a limit point of $\langle b_n \rangle$. Is $c+d$ necessarily a limit point of $\langle a_n + b_n \rangle$?

My Question:

When considering this question, do I have to sum over the same index or can the indices for the different subsequences differ? My intuition is that the subsequences must be summed over identical indices, in which case I believe that the following example serves as a counterexample:

Let $\langle a_n \rangle = (-1)^n, \ \langle b_n \rangle = (-1)^{n+1}$. Then summing over even and odd indices, I get $0 \ne 2$ or $-2$, which is the sum of their limit points. Have I done this correctly?

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Your argument is correct; but generally, the indices could differ. Take $a_n=b_n=(-1)^n$. $(a_n)$ has $1$ as a limit point and $(b_n)$ has $-1$ as a limit point but $(a_n+b_n)$ does not have $1+(-1)=0$ as a limit point. –  David Mitra Aug 14 '12 at 22:10

1 Answer 1

up vote 2 down vote accepted

Your example is fine. Both $\langle(-1)^n:n\in\Bbb N\rangle$ and $\langle(-1)^{n+1}:n\in\Bbb N\rangle$ have $1$ as a limit point, but $\langle(-1)^n+(-1)^{n+1}:n\in\Bbb N\rangle=\langle 0:n\in\Bbb N\rangle$ converges to $0$ and so does not have $1+1=2$ as a limit point.

It doesn’t matter what subsequence of $\langle a_n:n\in\Bbb N\rangle$ has $c$ as limit or what subsequence of $\langle b_n:n\in\Bbb N\rangle$ has $d$ as a limit; all that matters is whether some subsequence of $\langle a_n+b_n:n\in\Bbb N\rangle$ has $c+d$ as a limit. In your example that’s not the case, so yours is a genuine counterexample to the conjecture.

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@Rolando posted what looked like a proof of this conjecture, and then deleted it after you posted your answer. It looked something like this: Let $\epsilon >0$ be given. Since $a_n \to c$ and $b_n \to d$, there are $N_1, N_2$ such that for $l \ge N_1$, $k \ge N_2$ $|a_l - c| \le \epsilon/2$ and $|b_k - d| \le \epsilon/2$. Let $m = \max\{N_1, N_2\}$ and therefore we have for $n \ge m$ we have $|a_n - c + b_n - d| \le \epsilon$. Is the problem with his proof that he is not considering subsequences, and that he interpreted the limit points as limits? –  Mike Aug 14 '12 at 22:14
    
Wondering @BrianM.Scott –  Mike Aug 14 '12 at 22:16
    
@did: Thanks; fixed. –  Brian M. Scott Aug 14 '12 at 22:17
    
@jmi4 Yes, you're correct as to why the argument in your comment does not work. –  David Mitra Aug 14 '12 at 22:17
    
@jmi4: Rolando assumed that the original sequences were convergent. Some people use limit point of a sequence only to mean the unique limit point of a convergent sequence. They would probably use the term subsequential limit for what you’re calling a limit point of a sequence. –  Brian M. Scott Aug 14 '12 at 22:18

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