Real numbers/Sequences/Bounded increasing/Convergence/Decimal expansion/Section

Corollary

A bounded and monotone sequence in ${}\mathbb {R}$ converges.

Proof

Due to the condition, the sequence is increasing and bounded from above or decreasing and bounded from below. Because of fact, we have a Cauchy sequence which converges in ${}\mathbb {R}$.

$\Box$

This statement is also the reason that any decimal expansion defines a real number. An (infinite) decimal expansion

$a.a_{-1}a_{-2}a_{-3}\ldots$

with ${}a\in \mathbb {N}$ (we restrict to nonnegative numbers) and ${}a_{-n}\in \{0,\ldots ,9\}$ is just the sequence of rational numbers

$x_{0}:=a,\,x_{1}:=a+a_{-1}\cdot {\frac {1}{10}},\,x_{2}:=a+a_{-1}\cdot {\frac {1}{10}}+a_{-2}\cdot {\left({\frac {1}{10}}\right)}^{2},\,{\rm {{etc}.}}$

This sequence is increasing. It is also bounded, e.g. by ${}a+1$, so that it defines a Cauchy sequence and thus a real number.