Rational Sequence Increasing to Real Number

Theorem

Let $x \in \R$ be a real number.

Then there exists some increasing rational sequence that converges to $x$.

Proof

Let $\sequence {x_n}$ denote the sequence defined as:

$\forall n \in \N: x_n = \dfrac {\floor {n x} } n$

where $\floor {n x}$ denotes the floor of $n x$.

From Floor Function is Integer, $\floor {n x}$ is an integer.

Hence by definition of rational number, $\sequence {x_n}$ is a rational sequence.

$n x - 1 < \floor {n x} \le nx$

Thus:

$\dfrac {n x - 1} n < \dfrac {\floor {n x} } n \le x$

Further:

 $\displaystyle \lim_{n \mathop \to \infty} \frac {n x - 1} n$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \paren {x - \frac 1 n}$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} x - \lim_{n \mathop \to \infty} \frac 1 n$ Combined Sum Rule for Real Sequences $\displaystyle$ $=$ $\displaystyle x - 0$ Quotient Rule for Limits of Functions $\displaystyle$ $=$ $\displaystyle x$

From the Squeeze Theorem for Real Sequences:

$\displaystyle \lim_{n \mathop \to \infty} \frac {\floor {n x} } n = x$

From Peak Point Lemma, there is a monotone subsequence $\sequence {x_{n_k} }$ of $\sequence {x_n}$.

We have that $\sequence {x_n}$ is bounded above by $x$.

Hence $\sequence {x_{n_k} }$ is increasing.

From Limit of Subsequence equals Limit of Real Sequence, $\sequence {x_{n_k} }$ converges to $x$.

Hence the result.

$\blacksquare$