Problem with understanding Proof if two different subsequences have different limits then $(x_n)$ diverges

Question

As the title says I have problem understanding this proof. It seems ti ne that the proof is only about subsequences and not the main sequence itself. The proof I am referring to is from Two subsequences with different limits limplies **diverge** .

Problem

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

Answer

Pick a value of $c$. We wish to show that the sequence does not converge to $c$. Now, either $b\neq c$ or $a\neq c$ (or both). Without loss of generality, assume $a\neq c$.

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

(1) Our task, now, is to find some $s>N$ so that $|c−x_s|\ge\epsilon$.

(2) Since $\{x_{n_i}\}$ converges to $a$, let $s=n_i$, where $n_i$ is large enough so that $|a−x_{n_i}|<ϵ$ and $n_i\gt N$.

(3) 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, $$|c−x_s|>\frac{|c−a|}{2}=\epsilon$$

My understanding of the proof:

(1) This I understand. We want to show that $(x_n)$ diverges and this is the negation of the definition of limits of sequences of real numbers.

(2) Here $n_i$ is chosen such that it is large enough that $|a-x_{n_i}|<\epsilon$ but also $n_i>N$. I have troubles understanding why setting $s=n_i$.

(3) Here it is inferred from from $|a-x_{n_i}|<\epsilon$ that $|a-x_s|<\epsilon$ but $x_{n_i}$ and $x_s$ are members of two different sequences. Is it not possible that there is a $x_s$ such that $|a-x_s|\geq\epsilon$ even if $s=n_i>N$? If $x_s$ are members of the subsequence I don't know how this proof makes sense.

Edit: I don't buy that $|a−x_s|<ϵ$ as it says in (3). Just because there is a $M∈\mathbb{N}$ such that if $n_i≥M$ then $|a−x_{n_i}|<ϵ$ we can't infer from $n_i>N$ and $s=n_i$ that $|a−x_s|<ϵ$. Why? Because it's two different sequences we are talking about. There may be a number in $(x_n)$ even if $s>N$ such that $|a−x_s|>ϵ$ but |$a−x_{n_i}|<ϵ$.

Here I outline the proof using the method above to hopefully clearly show what I have problem understanding.

Suppose we have the alternating sequence $(x_n)=(1,-1,1...)$ and the subsequence $x_{n_i}=(1,1,1,...)$.

Start of proof that the sequence $(x_n)$ is divergent

Let $c\neq 1$, $\epsilon=|c-1|/2$ and $N\in\mathbb{N}$ be arbitrary. For any $\epsilon>0$ and $n_i\in\mathbb{N}$ we have that $|x_{n_i}-1|<\epsilon$. Let $n_i=s>N$, then $|c-1|=|c-x_s+x_s-1|\leq |c-x_s|+|x_s-1|<|c-x_s|+\epsilon$. Right here for me the proof breaks down. We want the proof to be about the main sequence so $x_s$ here are members of $(x_n)$. So $x_s$ is either $1$ or $-1$ so $|x_s-1|=0$ or $|x_s-1|=2$ depending if $s$ is even or odd.

Edit 2: Does the proof above specifically show that the subsequence can't have two different limits and thus $(x_n)$ is divergent?

Answer 1

Maybe this will be an easier proof.

Let $S= \{x_s\}$. Let $A \subset S$ and $A= \{x_{m_i}\}\subset S$ and suppose $A$ converges to $a$. Let $B\subset S$ and $B= \{x_{n_j}\}$ and suppose $B$ converges to $b \ne a$.

Let $\epsilon = \frac {b-a}4> 0$.

As $A$ converges to $a$, there will be a $K $ so that if $m_i > K$ we have $|a-x_{m_i}| < \epsilon$ and and $B$ converges to $b$, there is an $M$ so that if $n_j > M$ we have $|b-x_{n_j}|<\epsilon$

Claim: The is no $c\in \mathbb R$ and no $N\in \mathbb N$ where $s > N$ will imply $|c-x_s| < \epsilon$.

Pf: Pick any $c$ and any $N$. Let $W = \max(K, M, N)$.

Now $A$ and $B$ are infinite subsequences of $S$. So no matter how large $W$ we can always find a term $a_w \in A\subset S$ where $w > W\ge K$. And we can always find a term $a_v \in B\subset S$ where $v> W \ge M$.

(I think that might be the point you were missing. As $A$ and $B$ are infinite, no matter how far you go in the parent sequence $S$, there will always be terms of $A$ and there will always be terms of $B$ further along.)

(And the heart of the argument we can't have the terms that are in $A$ heading toward $a$ and the terms that are in $B$ heading toward $b$ while somehow having all the terms heading toward $c$. If they are all heading toward $c$ you cant have two sets of them heading in two different directions!)

Since $w > K$ and $x_w \in A$, we have $|a-x_w| < \epsilon$. If $|c-x_w| < \epsilon$ we can show that $|a-c|< 2\epsilon$. ($|a-c|=|a-x_w+x_w -c|< |a-x_w| +|x_w -c| < \epsilon + \epsilon = 2\epsilon$)

But we also have $v > W \ge M$ and $a_v \in B$, so we have $|b-x_w| < \epsilon$. If $|c-a_v| < \epsilon$ we can show that $|b-c| < 2\epsilon$ (ditto).

So if $c$ is so that both $|c-a_w| < \epsilon$ and $|c-a_v| < \epsilon$ we will have that both $|a-c| < 2\epsilon$ and $|b-c| < 2\epsilon$ and therefore $|a-b|=|a-c +c-b|\le |a-c|+|c-b|< 2\epsilon + 2\epsilon= 4 \epsilon = |b-a|$. Oops. That's impossible.

But as both $w, v > N$ and it is not possible so that both $|c-x_w| < \epsilon$ and $|c-x_w| < \epsilon$ and as $a_w\in A\subset S$ and $a_v\in B\subset S$ we have $x_w, x_v \in S$, it is not possible for ALL $s > N$ to imply $|c-x_s| < \epsilon$.

So it is not possible for there to be such an $N$ and $c$.

And if there is no such possible $c$ and $N$ we can not have $\lim x_n = c$.

======OLD POST BELOW (very long)

I think we may be getting confused by the notation of indexes and subindexes. The thing is it isn't actually very important but lets clear it up.

We have a sequence $a_1, a_2, a_3, a_4, .......$ The indexes of the sequence are $1,2,3,...$ and the cover the whole natural numbers.

We have subsequence $a_{something}, a_{somethingelse}, a_{athirdthing}....$ were the terms are all from the big sequence but not all of them. For example our subsequence migh be $a_3, a_5, a_9, a_{13}, a_{52}, .....$. We don't know which terms will or want be in the sequence. But we do know the fall in order. We can't have $a_3, a_5, a_9, a_{13}, a_{52}$ and then jump back to $a_7$ say.

Okay... so they way we reference this is to refer to the terms of the subsequence as $a_{n_1}, a_{n_2}, a_{n_3},....$ where $n_{k}< n_{k+1}$. In our example where the subsequence is $a_3, a_5, a_9, a_{13}, a_{52}$ we are referring to the indexes as $n_1 =3, n_2=5, n_3 =9, n_4=13$, etc and our subsequence is just refered to but $a_{n_1}, a_{n_2}, a_{n_3}, a_{n_4},.....$

Phew! This really isn't important but it will help.

Now we have two subsequence $a_{n_1},a_{n_2},....$ that converges to $a$ and another $a_{m_1}, a_{m_2},a_{m_3}.... $ that converges to $b \ne a$.

Now the proposal is we can't have the parent sequence converge to anything that is not $a$ or $b$ (and as everything is either not $a$ or not $b$ we can have the parent sequence converge to anything).

Okay? Here we go...

=====

Let $c\ne a$ then $\epsilon =\frac{|c-a|}2$ (which is the midpoint of $c$ and $a$) then we can find any $N$ so that if $k > N$ we will always have $|a_k -c| < \epsilon$.

Now the reason we won't be able to is because we have a subsequence ${a_{n_i}}$ where there is an $M$ where for all the $i > M$ we will have $a_{n_i}$ get close to $a$ which means we can't have all the $a_n$ where $n >M$ get close to $c$ which is far away from $a$.

So here's what the proof does: It says we want $N$, an arbitrary number, to be a "hurdle" for $a_n \to c$. We are going to prove that is impossible. We have $M$ which is a "hurdle" for the subsequence $a_{n_i} \to a$. [By "hurdle" I mean that for all $k > M$ we have $|a_{n_k} - a| < \epsilon$ and for all $s > N$ we have $|a_s - c| < \epsilon$.]

We want to find an $s$ where the hurdle for $|a_s-a| < \epsilon$ fails.

We do this by letting $i > M$ but letting $i$ be also large enough so that $n_i > N$. (We can do this by noting that as $n_1, n_2,n_3, ....$ are a subset of $1,2,3,4$ we always have $n_i \ge i$ so we let $i> \max(M,N)$ we will have $n_i \ge i > \max(M,N)\ge N$....but that wasn't important. As sequences go on forever we can always pick an index as far out as we need.)

So $n_i$ is a natural number. $a_{n_i}$ is the $i$th term of the subsequence $a_{n_i}$. And $n_i$ is a natural number. And $a_{n_i}$ is the $n_i$th term of the parent sequence $a_k$.

And $a_{n_i}$ is a term far out enough in the subsequence that converges to $a$ so that $|a-a_{n_i}| < \epsilon$. But $a_{n_i}$ is also a term of the parent sequence that is past the $n$th term.

Now the proof showed that if $|a-a_{n_i}| < \epsilon$ then $|c-a_{n_i}| > \epsilon$.

[because $|c−a|=|c−x_{n_i}+x_{n_i}−a|≤|c−x_{n_i}|+|a−x_{n_i}|<|c−x_{n_i}|+ϵ=|c−x_{n_i}|+\frac{|c−a|}{2}$ and therefore $|c−x_{n_i}|>\frac{|c−a|}{2}=\epsilon$]

That means.....

We can never find an $N$ where $n > N$ implies $|c-a_n| < \epsilon=\frac{|c−a|}{2}$ because for any $N$ we try we can always find an $n_i > N$ where $a_{n_i}$ is in the subsequence that converges to $a$ and where $|a-a_{n_i}| < \epislon$, and therefore $|c-a_{n_i}|> \epsilon$.

That's what the proof is saying. Is it clearer now?

Your questions.

2. Here ni is chosen such that it is large enough that |a−xni|<ϵ but also ni>N. I have troubles understanding why setting s=ni.

Ans: We wanted to find an $s$ so that $|c-a_s| > \epsilon$. We showed we can find an $n_i > N$ where $|c-a_{n_i}| > \epsilon$. SO we just assigned the value to $s$.

3)Here it is inferred from from |a−xni|<ϵ that |a−xs|<ϵ but xni and xs are members of two different sequences.

But $n_i = s$ and $a_{n_i}=a_s$ is a member of both sequences (but in different places in the sequences [but both are far enough that $a_{n_i}$ is past the hurdle for $|a-a_{n_i}|$ to be $< \epsilon$, and far enough that $n_i = s > N$])

3 cont) Is it not possible that there is a xs such that |a−xs|≥ϵ even if s=ni>N?

Of course. But we need to find an $N$ where ALL the values are such that $|c-x_s| < \epsilon$. We can't find an $N$ where ALL $s > N$ make that true, because there will always be some of then $n_i < N$ will be so that $|a-a_{n_i}| < \epsilon$ and therefore $|c-a_{n_i}| \ge \epsilon$. We can find some $s$ that work, but it will never be the case that ALL $s > N$ work.

EDIT: Why? Because it's two different sequences we are talking about.

But the subsequence is contained in the master sequence. No matter how far we go in the master sequence there will always be terms of the subsequence further along (because both are infinite). We can never find a point in the master where we will never have any more of the subsequence, and worse yet, we will always reach a point where all the further terms of the subsequence (which must occur) will all have to be too close to $a$.

Your question: Edit 2: Does the proof above specifically show that the subsequence can't have two different limits and thus (xn) is divergent?

Well, I don't understand your proof but if you have shown $|x_s -1| = 0$ if $s$ is odd$ and $|x_s-1| = 2$ if $s$ is even then, absolutely, this can never converge!