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I've already done a lot of searching around MSE for this. (In particular, If a sequence has two convergent subsequences with different limits, then it does not converge.)

The Cauchy criterion has not been covered yet.

I would like to show - without using the fact that a converging sequence has converging subsequences, in which both cases have the same limits - that if I have two subsequences with differing limits, the sequence does not converge.

That is, $\exists \epsilon > 0$ such that for all $M \in \mathbb{N}$, there is an $n \geq M$ such that $|x_n - x| \geq \epsilon$.

(I hope my negation above is correct.)

I am given two subsequences of $x_n$, namely $x_{n_i} \overset{i \to \infty}{\to}a$ and $x_{m_i} \overset{i \to \infty}{\to}b$, $a \neq b$.

Suppose, by way of contradiction, that $\lim\limits_{n \to \infty}x_n = x$.

I have the subsequence convergence, so something like $|x_{n_i} - a| < \text{?}$ and $|x_{m_i} - b| < \text{??}$.

I thought maybe to use a trick here: $$|x_n - x| = |x_{n}-x_{n_i}+x_{n_i}-a-x_{m_i}+x_{m_i}-b+b+a-x|$$ but I'm not sure how to proceed from here (triangle inequality isn't helpful).

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Clarinetist
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  • If the two limits are $l_1$ and $l_2$, let $\epsilon =\vert (l_1-l_2)\vert /2$ and negate the def of limit of a sequence. Drawing the picture will help. – Matematleta Jun 01 '16 at 15:03

3 Answers3

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As you know, the definition of $a_n \to c$ is $$\forall \epsilon>0\text{ } \exists N \in \Bbb{N} \text{ such that } s > N \implies |c - x_s| < \epsilon$$ The negation of this definition is simply, for any $c$, $$\exists\epsilon>0\text{ such that } \forall N \in \Bbb{N} \text{ } \exists s > N \text{ such that } |c - x_s| \ge \epsilon$$ And now we merely wish to show that this second statement applies to the series in question. Pick a value of $c$. We wish to show that the sequence does not converge to $c$. Now, either $b \ne c$ or $a \ne c$ (or both). Without loss of generality, assume $a \ne c$. Then let $\epsilon = |a-c|/2$. Now let $N$ be any natural number.

Our task, now, is to find some $s>N$ so that $|c - x_s| \ge \epsilon$. Since $(x_{n_i})$ converges to $a$, let $s = n_i$, where $n_i$ is large enough so that $|a-x_{n_i}| < \epsilon$ and $n_i > N$. Then $$|c - a| = |c - x_s + x_s - a| \le |c - x_s| + |a - x_s| < |c - x_s| + \epsilon = |c - x_s| + |c-a|/2$$ Therefore, $$|c - x_s| > |c-a|/2 = \epsilon$$ And we're done. We have shown that for any real number $c$, our sequence cannot converge to $c$. I hope this was helpful.

florence
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You don't need to consider two subsequences: it's easier to prove the statement in a different way.

Suppose the sequence $x_n$ converges to $a$. I'll prove that any subsequence converges to $a$.

A subsequence is given by an increasing function $i\mapsto n_i$. In particular, for every $N$ there is $I$ with $n_I>N$ (otherwise the range of $i\mapsto n_i$ would be finite) and so $n_i>N$ for every $i>I$.

So, let $\varepsilon>0$. By assumption, there exists $N$ with $|x_n-a|<\varepsilon$, for every $n>N$. If $n_i>N$ for every $i>I$, we have, in particular, that $|x_{n_i}-a|<\varepsilon$, for every $i>I$, thereby proving that the subsequence $x_{n_i}$ converges to $a$.

egreg
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Yes, your negation is correct (your notations are a little dangerous, though, $x_n$ being a series, $x$ a real and $x_{n_i}$ a subseries of $x_n$, but I will keep them here)

Let us consider $x \in \mathbb{R} \setminus a$

Let us pose $\epsilon = \frac{|a-x|}{2}$.

Let us consider $M \in \mathbb{N}$

With the limit of $x_{n_i}$ (and the definition of limit), we know there exists some $I \in \mathbb{N}$ such that $\forall i \geq I$, $|x_{n_i} - a| < \epsilon$

Let us pose $N = n_I $.

$|x_N-x|\geq |a-x| - \epsilon = \frac{|a-x|}{2} \geq \epsilon $

So $x_n$ cannot have a limit different than $a$.

Let us pose $\epsilon = \frac{a+b}{2}$

With the limit of $x_{m_i}$ (and the definition of limit), we know there exists some $I \in \mathbb{N}$ such that $\forall i \geq I$, $|x_{m_i} - b| < \epsilon$

Let us pose $N=m_I$

$|x_N-a| \geq |a-b| - \epsilon = \epsilon $

So $x_n$ cannot converge in $a$

Vincent
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