# 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. – Arthur Nov 30 '17 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 '17 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 '17 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). – Arthur Nov 30 '17 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 '17 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...
• 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. – David C. Ullrich Nov 30 '17 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... – David C. Ullrich Nov 30 '17 at 17:15
• 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 '17 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.