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).

  • $\begingroup$ 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. $\endgroup$ Jun 1 '16 at 15:03

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.


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 $a$.


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 $a$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.