# Show that if all convergent subsequences of a bounded sequence converge to $l$, the sequence itself must also converge to $l$.

Posted here for proof verification and corrections and tips!

Let $(a_n){^\infty_{n=0}}$ be a bounded sequence with the property that there exists $l$ such that if $(a_{n_j})^\infty_{j=1}$ is any convergent subsequence of $(a_n){^\infty_{n=0}}$ then its limit is $l$. Prove that $a_n\to l$ as $n\to\infty$.

Proof:

(a) $(a_n){^\infty_{n=0}}$ converges:

Suppose that all convergent subsequences of $(a_n){^\infty_{n=0}}$ converge to $l$, but $(a_n){^\infty_{n=0}}$ itself does not converge. As $(a_n){^\infty_{n=0}}$ is bounded, it cannot diverge to $\infty$. This means $(a_n){^\infty_{n=0}}$ must be alternating. We can look at the example $a_n=(-1)^n$. It is bounded by $[-1,1]$ but not convergent. However, for $a_n=(-1)^n$ there exist two convergent subsequences with different limits:

$a_{n_j}=(-1)^{j}$ with $j\in\{2n: n\in\mathbb{N}\}$ is constant and converges to $l_1=1.$

$a_{n_k}=(-1)^{k}$ with $k\in\{2n-1: n\in\mathbb{N}\}$ is constant and converges to $l_2=-1$.

This means not all the subsequences $a_{n_j}$ that are convergent, converge to the same $l$, so $(a_n){^\infty_{n=0}}$ cannot be alternating and must then be convergent.



(b) $(a_n){^\infty_{n=0}}$ has limit $l$.

Suppose that all convergent subsequences of $(a_n){^\infty_{n=0}}$ converge to $l$, but $(a_n){^\infty_{n=0}}$ converges to $x\neq l$. Then there exists a subsequence $(a_{n_j}){^\infty_{j=1}}=(-1)^{j}$ with $j\in\{n+1: n\in\mathbb{N}\}$ that converges to $x$, as it follows the original sequence, but starts at one element later. We assumed that $x\neq l$ but all convergent subsequences had limit $l$, but we just found a convergent subsequence that converges to $x\neq l$. This means $a_n){^\infty_{n=0}}$ must converge with its unique limit being $l$. $\tag*{$\Box$}$

• Alternating and diverging to $\infty$ are not the only two options (consider $\sin(n)$, for instance). Also, $a_n=n$ has the given property, with $l$ being for instance $\pi$: all converging subsequences (of which there are none) do converge to $\pi$. So some assumptions are missing. Nov 30, 2017 at 14:56
• I'm not understanding completely what assumptions are missing to make this correct. I do understand that alternating and diverging are not the only two options.
– Marc
Nov 30, 2017 at 15:09
• @Arthur what about: (1) $a_n$ is decreasing and bounded above $\implies$ there are no convergent subsequences. A contradiction. (2): $a_n$ is increasing and bounded above $\implies a_n$ is convergent. A contradiction. (3): $a_n$ is non-increasing and non-decreasing and bounded above $\implies a_n$ is alternating. The same arguments go for bounded below.
– Marc
Nov 30, 2017 at 15:09
• You seem to be using in your reasoning that $a_n$ is bounded, but this is not stated anywhere in the original statement. Maybe that's what's missing (it would certainly be enough to make me believe that it's true). Nov 30, 2017 at 15:20
• @Arthur you are totally correct! In fact, in the exercise it clearly stated that $a_n$ is bounded, but I just forgot to write that out.
– Marc
Nov 30, 2017 at 15:21

Suppose $a_n$ does not converge to $l$. By definition there exists $\epsilon>0$ such that for every $N$ there exists $n>N$ with $|a_n-l|>\epsilon$. Hence there is a subsequence $a_{n_j}$ with $$|a_{n_j}-l|>\epsilon$$for all $j$.

If the sequence $a_{n_j}$ were convergent we'd be done. There's no reason to think that it's convergent. But it's bounded, so...

• But it's bounded so... by definition, all subsequences are monotonically increasing, so the bounded, increasing subsequence is convergent? I just don't understand how we know the subsequence is bounded
– Marc
Nov 30, 2017 at 17:09
• The sequence $a_{n_j}$ is bounded because we're given that $a_n$ is bounded. No, that certainly does not imply that all subsequences are monotone. The sequence $(-1)^n$ is bounded. Nov 30, 2017 at 17:14
• What does it mean to say "$a_n$ is bounded"? It sounds like you don't know the definition - if not you don't have a chance here... Nov 30, 2017 at 17:15
• I'm sorry but... isn't part of the definition or a corollary of a subsequence, that it is strictly increasing?
– Marc
Nov 30, 2017 at 17:24
• And to say $a_n$ is bounded means that there exists an $l$ such that for all elements $a$ of $a_n$, $a<l$ holds.
– Marc
Nov 30, 2017 at 17:25

Assume $|a_n|\leq M$ for all $n\geq0$. If the sequence $(a_n)_{n\geq0}$ does not converge to $\ell$ then there is an $\epsilon_0>0$ such that there are infinitely many $n$ with $|a_n-\ell|\geq \epsilon_0$. These $a_n$ have are lying in the compact set $K:=[{-M},M]\setminus\>]l-\epsilon_0,l+\epsilon_0[\>$, hence have an accumulation point $\xi\in K$. It follows that there is a subsequence $k\mapsto a_{n_k}$ of the original sequence $(a_n)_{n\geq0}$ converging to $\xi$. Since $K$ does not contain the point $\ell$ we get a contradiction.