Theorem
Cauchy sequences are bounded.
Proof
Let \((a_n)\) be a Cauchy sequence and let \(\epsilon = 1\). By definition, this means that there exists a number \(N \in \mathbb{N}\) such that when \(n,m \geq N\)
$$
\begin{align*}
|a_n - a_m| < 1
\end{align*}
$$
Fix \(a_n\) such that \(n = N+1\). Since \(N+1 > N\), then the following holds for all \(m \geq N\)
$$
\begin{align*}
|a_m - a_{N+1}| < 1
\end{align*}
$$
By the reverse triangle inequality, then
$$
\begin{align*}
|a_m| - |a_{N+1}| \leq |a_m - a_{N+1}| < 1
\end{align*}
$$
Hence,
$$
\begin{align*}
|a_m| < 1 + |a_{N+1}|
\end{align*}
$$
From this, we can see that we have established a bound on all the terms of the sequence starting after the \(N\)th term. Now, define
$$
\begin{align*}
M = \max\{|a_1|, |a_2|, |a_3|, ..., |a_N|, |a_{N+1}|+1\}, \\
\end{align*}
$$
Hence, \(|a_n| \leq M\) for all \(n \in \mathbb{N}\). \(\blacksquare\)
References
- Introduction to Analysis, An, 4th edition by William Wade
- Lecture Notes by Professor Chun Kit Lai