Existence of Subdivision with Small Norm
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Theorem
Let $\closedint a b$ be a closed real interval.
Let $\epsilon > 0$ be a positive real number.
Then, there exists a finite subdivision $P$ of $\closedint a b$ such that:
- $\norm P < \epsilon$
where $\norm P$ denotes the norm of $P$.
Proof
By the Axiom of Archimedes, choose $N \in \N$ such that:
- $N > \dfrac {b - a} \epsilon > 0$
For each $k \in \set {0, 1, \dotsc, N - 1, N}$, define:
- $x_k = a + \dfrac k N \paren {b - a}$
We have:
| \(\ds x_0\) | \(=\) | \(\ds a + \frac 0 N \paren {b - a}\) | Definition of $x_k$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a\) | ||||||||||||
| \(\ds x_N\) | \(=\) | \(\ds a + \frac N N \paren {b - a}\) | Definition of $x_k$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds b\) | ||||||||||||
| \(\ds x_k - x_{k - 1}\) | \(=\) | \(\ds \paren {a + \frac k N \paren {b - a} } - \paren {a + \frac {k - 1} N \paren {b - a} }\) | Definition of $x_k$ | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {b - a} N\) | |||||||||||
| \(\ds \) | \(>\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x_{k - 1}\) | \(<\) | \(\ds x_k\) |
Let $P = \set {x_0, x_1, \dotsc, x_{N - 1}, x_N}$.
From:
- $x_0 = a$
- $x_N = b$
- $\forall k \in \set {1, \dotsc, N}: x_{k - 1} < x_k$
it follows that $P$ is a finite subdivision of $\closedint a b$.
Now:
| \(\ds \norm P\) | \(=\) | \(\ds \max \set {x_k - x_{k - 1} : k \in \set {1, \dotsc, N} }\) | Definition of Norm of Subdivision | |||||||||||
| \(\ds \) | \(=\) | \(\ds \max \set {\frac {b - a} N}\) | $\paren 1$ holds for all $k$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {b - a} N\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds \frac {b - a} {b - a} \epsilon\) | Definition of $N$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
$\blacksquare$