# 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 $m \ge N$:

\(\ds \size {a_m}\) | \(=\) | \(\ds \size {a_N + a_m - a_N}\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \size {a_N} + \size {a_m - a_N}\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \size {a_N} + 1\) |

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.

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 5$: Subsequences: Cauchy sequences: $\S 5.18$