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 $\sequence {x_n}$ denote the sequence defined as:
 * $\forall n \in \N : x_n = \dfrac {\ceiling {n x} } n$

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

From Ceiling Function is Integer, $\ceiling {n x}$ is an integer.

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

From Real Number is between Ceiling Functions:
 * $n x < \ceiling {n x} \le n x + 1$

Thus:
 * $x < \dfrac {\ceiling {n x} } n \le \dfrac {n x + 1} n$

Further:

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

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 below by $x$.

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

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

Hence the result.