# Cauchy Sequence is Bounded/Real Numbers

## 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 $m \ge N$:

 $\ds \size {a_m}$ $=$ $\ds \size {a_N + a_m - a_N}$ $\ds$ $\le$ $\ds \size {a_N} + \size {a_m - a_N}$ $\ds$ $\le$ $\ds \size {a_N} + 1$

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 $m \le N$:

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

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.

$\blacksquare$