Cauchy Sequence is Bounded/Real Numbers

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Theorem

Every Cauchy sequence in $\R$ is bounded.


Proof 1

Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.

Then there exists $N \in \N$ such that:

$\size {a_m - a_n} < 1$

for all $m, n \ge N$.


In particular, by the Triangle Inequality, for all $m \ge N$:

\(\displaystyle \size {a_m}\) \(=\) \(\displaystyle \size {a_N + a_m - a_N}\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {a_N} + \size {a_m - a_N}\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {a_N} + 1\)

So $\sequence {a_n}$ is bounded, as claimed.

$\blacksquare$


Proof 2

Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.

Then there exists $N \in \N$ such that:

$\size {a_m - a_n} < 1$

for all $m, n \ge N$.


Note that for $m \le N$:

\(\displaystyle \size {a_m}\) \(\le\) \(\displaystyle \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} }\)
\(\displaystyle \) \(<\) \(\displaystyle \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1\)
and for $m > N$:
\(\displaystyle \size {a_m}\) \(=\) \(\displaystyle \size {a_N + a_m - a_N}\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {a_N} + \size {a_m - a_N}\)
\(\displaystyle \) \(<\) \(\displaystyle \size {a_N} + 1\)
\(\displaystyle \) \(\le\) \(\displaystyle \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1\)

Hence for all $m \in \N$:

$\size {a_m} < \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1$


Therefore $\sequence {a_n}$ is bounded, as claimed.

$\blacksquare$


Sources