Subsequences and the Bolzano-Weierstrass Theorem

Exercise 2.5.5

Proof.
Let’s prove by contradiction. Suppose $(a_{n})$ does not converge to $a$. Since $(a_{n})$ does not converge to $a$, then there exists $\epsilon_{0} > 0$ such that for each $N \in \mathbb{N}$, we can find an $n \geq N$ where $|a_{n} - a| \geq \epsilon_{0}$. We can use this sequence to construct a subsequence that does not converge to $a$ using induction.

Base case: Let $N = 1$. There exists an index $n_{1} \geq 1$ such that $|a_{n_{1}} - a| \geq \epsilon_{0}$.

Inductive step: Assume we have found $n_{1} < n_{2} < n_{3} < \dots < n_{k}$ that satisfy $|a_{n_{k}} - a | \geq \epsilon_{0}$. To find the next term, let $N = n_{k} + 1$. Then there exists an index $n_{k+1} \geq n_{k} + 1$ such that $|a_{n_{k+1}} - a |\geq \epsilon_{0}$.

Because $n_{k+1} \geq n_{k}$, we have increasing indices and have constructed a subsequence $(a_{n_{k}})$ where $|a_{n_{k}} - a| \geq \epsilon_{0}$ for each $k \in \mathbb{N}$.

Since the original sequence $(a_{n})$ is bounded, the subsequence $(a_{n_{k}})$ is also bounded. By the Bolzano-Weierstrass Theorem, the bounded sequence $(a_{n_{k}})$ contains a convergent subsequence. Let’s call this convergent subsequence $(y_{m})$. Since $(y_{m})$ is a subsequence of $(a_{n_{k}})$, and $(a_{n_{k}})$ is a subsequence of $(a_{n})$, then $(y_{m})$ is a subsequence of $(a_{n})$. From the problem statement, every convergent subsequence of $(a_{n})$ must converge to $a$. Therefore, since $(y_{m})$ is a convergent subsequence of $(a_{n})$, then $(y_{m})$ must converge to $a$.

Since every term in $(y_{m})$ comes from $(a_{n_{k}})$, then every term satisfies $|y_{m} - a| \geq \epsilon_{0}$. If every term is at least $\epsilon_{0}$ distance away from $a$, the limit of $(y_{m})$ can not be $a$. We know this because by the Order Limit Theorem, the limit $L$ must satisfy $|L - a| \geq \epsilon_{0} > 0$. This is a contradiction: $(y_{m})$ converges to $a$ but is also bounded away from $a$. Therefore our initial assumption that $a_{n}$ does not converge to $a$ is false, and we conclude $(a_{n})$ converges to $a$. $\blacksquare$

Stephen Abbott, Understanding Analysis, 2nd edition.