User:Keith.U/Sandbox

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:
 * $\forall n \in \N : x_n = \dfrac {\left\lceil{ nx }\right\rceil} n$

where $\left\lceil{ \cdot }\right\rceil$ denotes the ceiling function.

We see immediately that $\left\langle{ x_n }\right\rangle$ is a rational sequence.

From Real Number Between Ceiling Functions:
 * $nx < \left\lceil{ nx }\right\rceil \leq nx + 1 $

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

Further:

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

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

Since $\left\langle{ x_n }\right\rangle$ is bounded below by $x$, $\left\langle{ x_{n_k} }\right\rangle$ must be decreasing.

Hence the result.