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:
| \(\ds \lim_{n \mathop \to \infty} \frac {nx + 1} n\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {x + \frac 1 n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} x + \lim_{n \mathop \to \infty} \frac 1 n\) | Combined Sum Rule for Real Sequences | |||||||||||
| \(\ds \) | \(=\) | \(\ds x + 0\) | Quotient Rule for Limits of Real Functions | |||||||||||
| \(\ds \) | \(=\) | \(\ds x\) |
Thus, from the Squeeze Theorem for Real Sequences:
- $\ds \lim_{n \mathop \to \infty} \frac {\ceiling {n x} } n = x$
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.
$\blacksquare$