# 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:

\(\displaystyle \lim_{n \mathop \to \infty} \frac {nx + 1} n\) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \paren {x + \frac 1 n}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} x + \lim_{n \mathop \to \infty} \frac 1 n\) | Combined Sum Rule for Real Sequences | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x + 0\) | Quotient Rule for Limits of Functions | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x\) |

Thus, from the Squeeze Theorem for Real Sequences:

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

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

Hence the result.

$\blacksquare$