Two subsequences with different limits limplies **diverge**

calculuslimitssequences-and-series

Problem_

Prove that if two subsequences of a sequence $\{a_n\}$ have different limits $a \ne b$, then $\{a_n\}$ diverges.

In fact, I've seen several proofs of the question. But, unfortunately, I cannot understand some of those solutions. One proof that I saw the most is:

Pick a value of $c$. We wish to show that the sequence does not converge to $c$. Now, either $b≠c$ or $a≠c$ (or both). (1) By WLOG, assume $a≠c$.

Then let $ϵ=\frac{|a−c|}{2}$. Now let N be any natural number.

Our task, now, is to find some $s>N$ so that $|c−x_s|\geϵ$. Since $\{x_{n_i}\}$ converges to $a$, (2) let $s=n_i$, where $n_i$ is large enough so that $|a−x_{n_i}|<ϵ$ and $n_i\gt N$.

Then
$$|c−a|=|c−x_s+x_s−a|≤|c−x_s|+|a−x_s|<|c−x_s|+ϵ=|c−x_s|+\frac{|c−a|}{2}$$
Therefore,
$|c−x_s|>\frac{|c−a|}{2}=ϵ$

(3) And we're done.

See more at the following site:

[1] Two subsequences with different limits $\implies$ not convergent

[2] If a sequence has two convergent subsequences with different limits, then it does not converge

Here are my questions:

Questions_

(1) How can 'WLOG' be applied in this case?

(2) Can we set $s$ as $n_i$ without any other conditions?

(3) The solution only analyzed the subsequence converges to $a$. How about the other one(converges to $b$)? Is it covered by WLOG? Then, how?

Thanks for giving me ideas.

Best Answer

  1. "WLOG" means "without loss of generality." In this case, we know either $b \neq c$ or $a \neq c$. But there really is no distinction between $a$ and $b$ in this problem - they are both limits of some subsequence of $\{a_n\}$, but there is no other distinction between them. Thus we could safely "rename" them if we wanted to in order to have $a \neq c$. An alternative way of looking at it is that we could go through the proof under the assumption $a \neq c$, then go through the analogous proof again under the assumption that $b \neq c$, and come to the same conclusion in either case. Saying "WLOG, $a \neq c$" allows us to avoid writing essentially the same proof twice for the two different cases.
  2. I'm not sure exactly what you're asking here. But we know that since a subsequence of $\{a_n\}$ converges to $a$, there must be some $M$ such that for all $n_i>M$, $|a_{n_i}-a|<\epsilon$, where $n_i$ is the sequence of indices for the subsequence converging to $a$. Now we want $n_i>N$, and since we know the above holds for all $n_i>M$, we can certainly find an $n_i>N$ for which it holds. If you'd like, take $n_i$ to be the smallest $n_i$ greater than $\max(M,N)$.
  3. See point 1.
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