Paving Lemma

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ be an open subset of the Euclidean space $\R^m$ or the set of complex numbers $\C$.

Let $\gamma: \left[{a \,.\,.\, b}\right] \to S$ be a path in $S$.


Then there exists $K \in \R_{>0}$ such that:

For all $\epsilon \in \left({0 \,.\,.\, K}\right)$, there exists a normal subdivision $\left\{{x_0, x_1, \ldots, x_{n-1}, x_n}\right\}$ of the closed interval $\left[{a \,.\,.\, b}\right]$ such that:
$\displaystyle \bigcup_{i \mathop = 0}^n B_\epsilon \left({\gamma \left({x_i}\right) }\right) \subseteq S$
and for all $i \in \left\{ {0, 1, \ldots, n - 1}\right\}$:
$\gamma \left({\left[{x_i \,.\,.\, x_{i + 1} }\right] }\right) \subseteq B_\epsilon \left({\gamma \left({x_i}\right) }\right)$


Here, $B_\epsilon \left({\gamma \left({x_i}\right) }\right)$ denotes the open ball of $\gamma \left({x_i}\right)$ with radius $\epsilon$.


Proof

Finding the constant

First, suppose that $S$ is a subset of $\R^m$.

From Closed Real Interval is Compact, it follows that $\left[{a \,.\,.\, b}\right]$ is compact.

Then Continuous Image of Compact Space is Compact shows that $\gamma \left({\left[{x_i \,.\,.\, x_{i + 1} }\right] }\right)$ is compact.

From the Heine-Borel Theorem, it follows that $\gamma \left({\left[{x_i \,.\,.\, x_{i + 1} }\right] }\right)$ is bounded and closed.


Suppose that $S \ne \R^m$.

As $\R^m \setminus S$ is closed, it follows from Distance between Closed Sets in Euclidean Space that:

$d \left({\gamma \left({\left[{a \,.\,.\, b}\right] }\right), \R^m \setminus S}\right) > 0$

Put $K = d \left({\gamma \left({\left[{a \,.\,.\, b}\right] }\right), \R^m \setminus S}\right)$, the distance between $\gamma \left({\left[{a \,.\,.\, b}\right] }\right)$ and $\R^m \setminus S$.

If instead $S = \R^m$, then $K$ may be any strictly positive real number.


Finding the subdivision

Let $\epsilon \in \left({0 \,.\,.\, K}\right)$.

From the Heine-Cantor Theorem, it follows that $\gamma$ is uniformly continuous.

Then there exists $\delta \in \R_{>0}$ such that:

$\forall y, z \in \left[{a \,.\,.\, b}\right]: \left\vert{y - z}\right\vert < \delta$ implies $d \left({\gamma \left({y}\right),\gamma \left({z}\right)}\right) < \epsilon$

where $d$ denotes the Euclidean metric.

Choose $n \in \N$ such that $\dfrac {b - a} n < \delta$, and put $x_i = a + i \dfrac {b - a} n$ for all $i \in \left\{ {0, 1, \ldots, n}\right\}$.

Then $\left\{ {x_0, x_1, \ldots, x_n}\right\}$ is a normal subdivision of $\left[{a \,.\,.\, b}\right]$.

It follows that $B_\epsilon \left({\gamma \left({x_i}\right) }\right)$ and $\R^m \setminus S$ are disjoint, as either:

$\epsilon < d \left({\gamma \left({\left[{a \,.\,.\, b}\right] }\right), \R^m \setminus S}\right)$

or:

$\R \setminus S = \varnothing$

Then:

$\displaystyle \bigcup_{i \mathop = 0}^n B_\epsilon \left({\gamma \left({x_i}\right) }\right) \subseteq S$

Let $x \in \left[{x_i \,.\,.\, x_{i+1} }\right]$.

Then:

$\left\vert{x - x_i}\right\vert \le \dfrac{b - a} n < \delta$

As the uniorm continuity condition applies, it follows that:

$\gamma \left({\left[{x_i \,.\,.\, x_{i + 1} }\right] }\right) \subseteq B_\epsilon \left({\gamma \left({x_i}\right) }\right)$

$\Box$


Complex plane case

Suppose that $S$ is a subset of $\C$.

This theorem only uses the properties of $\C$ that depends on the metric.

From Complex Plane is Metric Space, it follows that $\C$ and $\R^2$ are homeomorphic.

It follows that the proof that applies for $\R^2$ also applies for $\C$.

$\blacksquare$


Sources