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 $\left\langle{x_n}\right\rangle$ denote the sequence defined as:
 * $\forall n \in \N: x_n = \dfrac {\left\lfloor{n x}\right\rfloor} n$

where $\left\lfloor{n x}\right\rfloor$ denotes the floor of $n x$.

From Floor Function is Integer, $\left\lfloor{n x}\right\rfloor$ is an integer.

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

From Real Number is between Floor Functions:
 * $n x - 1 < \left\lfloor{n x}\right\rfloor \le nx $

Thus:
 * $\dfrac {n x - 1} n < \dfrac {\left\lfloor{n x}\right\rfloor} n \le x$

Further:

From the Squeeze Theorem for Sequences of Real Numbers:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac {\left\lfloor{n x}\right\rfloor} n = x$

From Peak Point Lemma, there is a monotone subsequence $\left\langle{x_{n_k}}\right\rangle$ of $\left\langle{x_n}\right\rangle$.

We have that $\left\langle{x_n}\right\rangle$ is bounded above by $x$.

Hence $\left\langle{x_{n_k}}\right\rangle$ is increasing.

From Limit of Real Subsequence equals Limit of Real Sequence, $\left\langle{x_{n_k}}\right\rangle$ converges to $x$.

Hence the result.