51
$\begingroup$

Is the following true?
Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.

I guess that it is true but I am not sure how to prove this.

$\endgroup$
47
$\begingroup$

True. If not, there exists an $\epsilon > 0$, such that for all $k$, there exists an $n_k > k$ satisfying $|x_{n_k}−x| \ge \epsilon$ since if there is some $k$ which doesn't have such $n_k$, then we can take it as $N$, so $x_n$ converges to $x$. The subsequence $x_{n_{k}}$ does not have any subsequence converging to $x$.

| cite | improve this answer | |
$\endgroup$
  • 7
    $\begingroup$ This works in an arbitrary topological space: "If not, there exists a neighborhood $U$ of $x$ such that for all $k$, there is an $n_k > k$ such that $x_{n_k}\notin U$. Otherwise, $\{x_n\}$ is eventually in all neighborhoods of $x$, which contradicts $x_n\not\to x$. Then, $\{x_{n_k}\}$ has no subsequences converging to $x$, since it is never in $U$." $\endgroup$ – Michael L. Aug 22 '17 at 11:08
  • 2
    $\begingroup$ Even better, the result can be extended to nets. $\endgroup$ – Michael L. Aug 22 '17 at 11:13
3
$\begingroup$

proof by Contradiction

Suppose that $\{X_n\}$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists \hspace{.2cm}n=n(N) : n>N~~~and ~~~ |X_n -\ell|>\varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that $$n_1>N_1 ~~~and ~~~ |X_{n_1} -\ell|>\varepsilon_0 $$ Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1>N_{k+1}}$ such that,

$$ |X_{ n_{k+1}} -\ell|>\varepsilon_0 $$

It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$ since $$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$

However, $$\forall k,~~ |X_{ n_{k}} -\ell|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $ but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a congering subsequence of $\{X_n\}_n$

By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have

$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{CONTRADICTION}$$

Note that $$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0$$ Since $$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ BW seems overkill here... $\endgroup$ – Asier Calbet Feb 25 '19 at 17:19
  • $\begingroup$ (+1), very comprehensive answer. :) By the way, do you know any direct proof for this? If, yes, please take a look at my question here. $\endgroup$ – Hosein Rahnama Feb 19 at 12:04
2
$\begingroup$

This is much easier to prove than the other answers here might suggest, if you remember the handy little fact (given, for instance, in Baby Rudin but which aught to be more well-known) that a sequence $x_n$ does not converge to $x$ if, and only if, there exists $\epsilon>0$ such that $|{x_n-x}| \geq \epsilon$ for infinitely many $n$.

On the one hand, if $x_n \to x$ then this fact gives that any subsequence $(x_{n_i}) \subset (x_i)$ can also have finitely many terms a distance $\epsilon$ from $x$, and therefore also converges to $x$. In particular, if $x_n$ converges to $x$ then every subsequence has a subsequence converging to $x$, namely itself

On the other hand, if $x_n \not\to x$ then there exist infinitely many $n$ such that $|x_n-x| \geq \epsilon$ and the subsequence of these terms has no subsequence converging to $x$ as any subsequence has all terms a distance of at least $\epsilon$ away from $x$

Note, however, that it is necessary that the sub-subsequnces converge to the same limit: for instance, any subsequence of the alternating sequence $0,1,0,1,0,1,\ldots$ has a convergent subsequence by the pigeonhole principle, but such sub-subsequences can converge to either 0 or 1

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Assume $(x_n)$ doesn't converge to $x$. In case $1, (x_n)$ isn't bounded, thus has at least one sub-sequence going to either $\infty$ or $-\infty.$ This subsequence clearly doesn't have any subsequence converging to $x$ (thus a contradiction).

In case $2, (x_n)$ is bounded. If $(x_n)$ converges to a finite real limit different from $x$ then you get an immediate contradiction to the property. Otherwise $(x_n)$ is divergent and bounded thus has $\liminf$ and $\limsup.$ These two converge, and have a sub-sequence which converges to $x$, thus must converge to $x$ themselves (otherwise you get a similar contradiction). And this means that : $$\liminf=\limsup=0 \implies \lim(x_n)=0$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ correction : by 0 - I meant x. $\endgroup$ – nimrod de la vega Aug 6 '17 at 17:05

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.