Convergence of a Bounded Sequence is determined by Convergent Subsequences

Last updated Nov 1, 2022

Let $(a_{n}) \subset \mathbb{R}$ be a Bounded Sequence. Then $a_{n} \to a \in \mathbb{R}$ iff every convergent Subsequence of $(a_{n})$ converges to $a$.

($\Rightarrow$) Suppose $a_{n} \to a$. Let $(a_{n_{k}})$ be a convergent subsequence of $(a_{n})$. Because Every Subsequence of a Convergent Sequence converges to the same Limit, we know $a_{n_{k}} \to a$.

($\Leftarrow$) Suppose every convergent subsequence of $(a_{n})$ converges to $a \in \mathbb{R}$. First we show that such a subsequence exists.

Let $M \in \mathbb{R}$ be s.t. $|a_{n}| \leq M$ $\forall n \in \mathbb{N}$. By the Bolzano-Weierstrass Theorem, we know there exists some convergent subsequence.

Next we show that $(a_{n}) \to a$. Suppose not. Then there exists an $\epsilon > 0$ s.t. $\forall N \in \mathbb{N}$, $n \geq N$ s.t. $|a_{n} - a| \geq \epsilon$. Let $n_{1}$ be such an element for $N = 1$. Next let $n_{2}$ be such an element for $N = 2$. Continue in this way to get a subsequence $(a_{n_{k}})$ s.t.

$$|a_{n_{k}} - a| \geq \epsilon \text{ for all }k \in \mathbb{N}$$ Because ${a_{n_{k}}} \subset {a_{n}}$, it has the same bound of $M$. Thus by the Bolzano-Weierstrass Theorem, $(a_{n_{k}})$ has a convergent subsequence. This subsequence is also a subsequence of $(a_{n})$ so it must also converge to $a$ by our given. If we denote this sub-subsequence $(a_{n_{k_{l}}})$, we have that there exists an $L \in \mathbb{N}$ s.t.

$$|a_{n_{k_{l}}} - a| < \epsilon \text{ for all }l \in \mathbb{N}$$ But this contradicts our construction of $(a_{n_{k}})$. Therefore $(a_{n}) \to a$. $\blacksquare$

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