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

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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$}$