Rational Sequence Increasing to Real Number
Theorem
Let $x \in \R$ be a real number.
Then there exists some increasing rational sequence that converges to $x$.
Proof
Let $\sequence {x_n}$ denote the sequence defined as:
- $\forall n \in \N: x_n = \dfrac {\floor {n x} } n$
where $\floor {n x}$ denotes the floor of $n x$.
From Floor Function is Integer, $\floor {n x}$ is an integer.
Hence by definition of rational number, $\sequence {x_n}$ is a rational sequence.
From Real Number is between Floor Functions:
- $n x - 1 < \floor {n x} \le n x$
Thus:
- $\dfrac {n x - 1} n < \dfrac {\floor {n x} } n \le x$
Further:
\(\ds \lim_{n \mathop \to \infty} \frac {n x - 1} n\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {x - \frac 1 n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} x - \lim_{n \mathop \to \infty} \frac 1 n\) | Combined Sum Rule for Real Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds x - 0\) | Quotient Rule for Limits of Real Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) |
From the Squeeze Theorem for Real Sequences:
- $\ds \lim_{n \mathop \to \infty} \frac {\floor {n x} } n = x$
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 above by $x$.
Hence $\sequence {x_{n_k} }$ is increasing.
From Limit of Subsequence equals Limit of Real Sequence, $\sequence {x_{n_k} }$ converges to $x$.
Hence the result.
$\blacksquare$