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 $m, n \ge N$.


So for all $m \ge N$, we have:

\(\ds \size {a_m}\) \(=\) \(\ds \size {a_N + a_m - a_N}\)
\(\ds \) \(\le\) \(\ds \size {a_N} + \size {a_m - a_N}\) Triangle Inequality
\(\ds \) \(\le\) \(\ds \size {a_N} + 1\)

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

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