If two subsequences converge to different values, prove directly the sequence itself diverges

Question

Let $\{x_n\}$ be a sequence. Suppose that there are two convergent subsequences $\{x_{n_{\Large{i}}} \}$ and $\{x_{m_{\Large{i}}} \}$. Suppose that $\lim\limits_{i\to\infty} x_{n_{\Large{i}}} = a$ and $\lim\limits_{i\to\infty} x_{m_{\Large{i}}} = b$, where $a \ne b$. Prove that $\{x_n\}$ is not convergent.

I have to prove this without using the idea that subsequences are convergent if the sequence is convergent and their limits are equal. I have no idea where to go with this problem and have been stuck on it for quite some time now.

score 4 · Answer 1 · answered Jun 30 '13 at 21:56

4

Let $\epsilon=\dfrac{|b-a|}3,$ and note that $\epsilon>0$.

Now, show that for all $N,$ there exist $m,n\ge N$ such that $|x_n-x_m|\ge\epsilon$. Thus, the sequence is not Cauchy, and so not convergent. Hint: If $|x_{n_i}-a|\le\epsilon$ and $|x_{m_i}-b|\le\epsilon,$ what can we say about $|x_{n_i}-x_{m_i}|$ in light of the fact that $|b-a|=3\epsilon$?

answered Jun 30 '13 at 21:56

Cameron Buie

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This seems to be where I keep getting stuck. I get confused on how to show that such an m and n exist. – user72195 Jun 30 '13 at 21:57
I have rephrased my post slightly to make the hint a bit clearer. Does that help? – Cameron Buie Jun 30 '13 at 22:00
yes it does, thank you. – user72195 Jun 30 '13 at 22:01
although, how do I create the relationship between the two inequalities? – user72195 Jun 30 '13 at 22:05
I recommend that you apply the triangle inequality to $|b-a|,$ after rewriting "$b-a$" in an equivalent way. Think on it for a bit, and let me know if you're still stuck. – Cameron Buie Jun 30 '13 at 22:10

If two subsequences converge to different values, prove directly the sequence itself diverges

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