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

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.

• Can you try to write in a clear manner whats your trouble with the argument? – aram Nov 4 '14 at 4:56
• 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 $x$ . – John Nov 4 '14 at 5:01
• 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 – user189731 Nov 4 '14 at 5:03
• 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? – bof Nov 4 '14 at 5:21

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 $x$?

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 $\varepsilon'$

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

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 $x$.