2
$\begingroup$

If every subsequence has a further subsequence that converges to $x$, then the sequence converges to $x$

In the proof, we assume the contrary, then there is some $\epsilon$ such that $|x_{n} -x|> \epsilon$, $\forall n>N.$ Then we can find some subsequence which also doesn't converge to $x$.

But then can't we still find a further subsequence which may converge to $x$? There certainly are further subsequence that don't converge, but it doesn't mean that all further subsequences don't converge?

Please clarify the last part.

$\endgroup$
4
  • $\begingroup$ Can you try to write in a clear manner whats your trouble with the argument? $\endgroup$ – aram Nov 4 '14 at 4:56
  • $\begingroup$ Is this your orignial statment? If there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ . $\endgroup$ – John Nov 4 '14 at 5:01
  • $\begingroup$ What I meant was we can find a subsequence {$x_{n_k} $} st |$x_{n_k} $ - x| > $epsilon$. But why does this imply that there's no further subsequence which converges to x? A sequence diverging from x doesn't mean that all subsequences diverge as well $\endgroup$ – user189731 Nov 4 '14 at 5:03
  • 1
    $\begingroup$ The statement in the title of your question is false. Perhaps you meant to say, if every subsequence has a further subsequence that converges to x then the sequence converges to x? $\endgroup$ – bof Nov 4 '14 at 5:21
7
$\begingroup$

My experience was that this statement is easiest to prove by contraposition. In this case you want to prove that if $x_n$ does not converge to $x$ then there exists a subsequence $x_{n_k}$ such that all further subsequences $x_{n_{k_\ell}}$ do not converge to $x$.

If we write the definition of $x_n \not \to x$, we have

$$(\exists \varepsilon > 0)(\forall N \in \mathbb{N})(\exists n \geq N) \, |x_n - x| > \varepsilon.$$

This definition allows us to extract a subsequence which is in fact bounded away from $x$: we fix such a $\varepsilon$, then we apply the universal for each $N \in \mathbb{N}$, which must necessarily give us countably many $n$ with the desired property. Why does this subsequence not have a further subsequence which converges to $x$?

$\endgroup$
1
$\begingroup$

Al right I think I see whats your problem. When you asume $x_n \not\rightarrow x$, what you are saying is:

There exist some $\varepsilon > 0$ such that for all $n \in \mathbb{N}$ (you fix an $n$, show an $n_k$, fix another $n$ show another $n_k$, etc) we have some $n_k >n$ such that $|x_{n_k} - x| \geq \varepsilon$. In simpler terms what you get is a subsequence such that $|x_{n_k} -x| \geq \varepsilon$ for all $n_k$ and this subsequence can't converge by definition.

Remembering the definition of convergence: $x_n \rightarrow x$ means that given an arbitrary $\varepsilon' > 0$, there exists some $N$ such that for all $n \geq N$, $|x_n -x |< \varepsilon'$.

Now if you take $\varepsilon' < \varepsilon$ you would see that $x_{n_k} \not\rightarrow x$ because every $x_{n_k}$ is at distance $\varepsilon$ from $x$ wich is greater than $\varepsilon'$

$\endgroup$
2
  • $\begingroup$ What about the countered ample where $x{_n}$ = 1,0,1,0,1,0... Then $x_{n_k}$ = 1,1,1,... converges to 1. $\endgroup$ – user189731 Nov 4 '14 at 5:21
  • 1
    $\begingroup$ @user189731 This sequence doesnt comply with the hypothesis of the theorem you were talking about. $\endgroup$ – aram Nov 4 '14 at 5:23
1
$\begingroup$

For a sequence $x_n$ define $\alpha = \limsup x_n$ and $ \beta = \liminf x_n $, which always exist, and for simplicity assume that they are finite. Consider the subsequence that converges to $\alpha$, denote it by $x_{n_k}$. Then this subsequence has a further subsequence that converges to $x$ by definition. But since the limits are unique we must have $x=\alpha$. Similarly, $\beta = x$. Thus

$$\liminf x_n = \limsup x_n = x $$

And so the sequence converges to $x$.

$\endgroup$

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.