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We have the following elementary result on real sequences.

Any convergent sequence is bounded.

This is basically the proof given in my notes:

Suppose that $a_n \to a \in \mathbb{R}$. Now choose $\epsilon = 1$. From the $\epsilon-N$ definition of convergence, $\exists N \in \mathbb{N}: n > N \implies |a_n - a| < 1$. The triangle inequality gives us $|a_n| \leq |a| + 1$ for $n > N$. Therefore, choosing $C = \sup \{a_n | n \leq N\} \cup \{|a| + 1\}$ we obtain $|a_n| < C$, so $a_n$ is bounded. $\Box$

It seems to me that it is possible that, for $n < N$, $\sup \{a_n\} < -a_n$ and therefore$ \sup\{a_n\} < |a_n|$. This will be the case if, for $n < N$, $a_n$ is very negative but bounded above by a relatively small positive quantity. Therefore, we actually need $C = \sup \{|a_n| | n \leq N\} \cup \{|a| + 1\}$ for this proof to work.

Am I correct?

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    $\begingroup$ Yes, you are right. $\endgroup$ – Mark Apr 17 at 21:32
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Yes, you are correct. Everything in their proof is correct until they choose $C$.

$|a_n|<|a|+1$ for all $n>N$ and $|a_n|\leq\sup\{|a_n|:n\leq N\}$, so $|a_n|\leq \sup(\{|a_n|:n\leq N\}\cup\{|a|+1\})$ for all $n$.

Take any sequence $(a_n)$ such that $a_n<0$ and $|a_n|>|a|+1$ for all $n\leq N$, and it will not be bounded in the way that their proof asserts.

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