Prove that if $({a_n} )$ is a sequence with two convergent subsequences $\{a_{n_k}\}$ and $\{a_{m_k}\}$

Question

Prove that if $({a_n} )$ is a sequence with two convergent subsequences $\{a_{n_k}\}$ and $\{a_{m_k}\}$ such that $\lim_{k\to\infty}\{a_{n_k}\} \neq \lim_{k\to\infty}\{a_{m_k}\}$, then $({a_n})$ does not converge.

I am not sure if I am approaching the question correctly but, I have two limits set up to two different answers, $|\{a_{n_k}\} - L|<\epsilon/2$ and $|\{a_{m_k}\} - M|<\epsilon/2$ because I thought if I could somehow show these can not be manipulated to equal $\epsilon$ then their limits could not be equal to each other and therefore $(a_n)$ would not converge. I'm not sure if I can set the limits equal to an answer without proving something first or if I am actually going the right way about this. I am a little confused by the question in general, any help would be much appreciated.

Show that if $(a_n)$ converges ($\lim a_n= L$) then any subsequence of $(a_n)$ converges to L. — PAM1499, Sep 26 '20 at 20:52
The very text of the question says that $a^k_n$ and $a^k_m$ are convergent subsequences. So yes, by hypothesis you are allowed to set their limits equal to an answer. I think that the proof is by contradiction. Suppose $(a_n)$ converges then show that this cannot be possible. To show that this cannot be possible, I believe you'll have to use the definition of limit. — Sigma, Sep 26 '20 at 21:02
Does this answer your question? If a sequence has two convergent subsequences with different limits, then it does not converge — Martin R, Sep 26 '20 at 21:10
@MartinR, yes that is what I meant, I got confused with notation but I fixed it. — Sam, Sep 26 '20 at 21:51
@Sam: It is enough to show the sequence is not Cauchy. Suppose $L$ and $M$ are the limits of the convergent subsequences and that $L\neq M$. Let $\varepsilon=\frac{|M-L|}{2}$. For any $K$, there are $n_K,m_K>K$ such that $|a_{n_K|}-a_{m_K}|\geq\varepsilon$. — Mittens, Sep 26 '20 at 22:43

Mittens · Answer 1 · 2020-09-26T23:27:43.520

Recall that a sequence $\{a_n:n\in\mathbb{N}\}$ is Cauchy if for any $\varepsilon>0$, there is $N$ such that $n,m\geq N$ then $|a_n-a_m|<\varepsilon$.

We have the following well known result:

Proposition: If a sequence $\{a_n:n\in\mathbb{N}\}$ is convergent, then it is Cauchy.

Here is a short proof: Suppose $a_n\xrightarrow{n\rightarrow\infty}a$. Then, for any $\varepsilon>0$, there is $N$ such that $n\geq N$ implies $|a_n-a|<\varepsilon/2$. Consequently, if $n,m\geq N$, $|a_n-a_m|\leq |a_n-a|+|a-a_m|<\varepsilon$.

Notice that the negation of being Cauchy is: $\{a_n\}$ is not Cauchy if there is $\varepsilon>0$ such that for any $N$, there exist $m,m\geq N$ such that $|a_n-a_m|\geq\varepsilon$.

Under the condition of the original OP, we show $\{a_n\}$ is not Cauchy:

Suppose $a_{n_k}\xrightarrow{k\rightarrow\infty}L$, $a_{m_k}\xrightarrow{k\rightarrow\infty}M$ and $M\neq L$. Let $$\varepsilon:=\frac{|L-M|}{4}$$ By assumption, there is $K_0$ such the $k\geq K_0$ implies that $$|a_{n_k}-L|<\varepsilon$$ and $$|a_{m_k}-M|<\varepsilon$$

Now, given any integer $N$, choose $k\geq \max(K_0,N)$. Then, $n_k\geq k\geq N$ and $m_k\geq k\geq N$, and $$ |a_{n_k}-a_{m_k}|\geq \varepsilon $$ (without loss of generality assume $L<M$ and draw a picture to see what is going on. A rigorous argument is based on the triangle inequality).

This means that $\{a_n\}$ is an a Cauchy sequence and so, $\{a_n\}$ is not convergent.

user · Answer 2 · 2020-09-26T21:08:37.590

0

Just assume wlog $L>M$ and $\varepsilon \le \frac{L-M}{3}$ then since we can find $n_0$ such that for any $n\ge n_0$

$$|a^k_n - L|<\varepsilon \quad \land \quad |a^k_m - M|<\varepsilon$$

which are disjointed intervals

$$\frac{-L+4M}{3}<a^k_m<\frac{L+2M}{3}<\frac{2L+M}{3}<a^k_n <\frac{4L-M}{3} $$

then $a_n$ doesn't converge.

edited Sep 26 '20 at 21:08

answered Sep 26 '20 at 21:03

user

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Prove that if $({a_n} )$ is a sequence with two convergent subsequences $\{a_{n_k}\}$ and $\{a_{m_k}\}$

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