Rational Sequence Decreasing to Real Number

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

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

Proof
Let $\left\langle{x_n}\right\rangle$ denote the sequence defined as:
 * $\forall n \in \N : x_n = \dfrac {\left\lceil{n x}\right\rceil} n$

where $\left\lceil{n x}\right\rceil$ denotes the ceiling of $n x$.

From Ceiling Function is Integer, $\left\lceil{n x}\right\rceil$ is an integer.

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

From Real Number is between Ceiling Functions:
 * $n x < \left\lceil{n x}\right\rceil \le n x + 1$

Thus:
 * $x < \dfrac{\left\lceil{n x}\right\rceil} n \le \dfrac {n x + 1} n$

Further:

Thus, from the Squeeze Theorem for Sequences of Real Numbers:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac {\left\lceil{n x}\right\rceil} n = x$

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 below by $x$.

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

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

Hence the result.