Paving Lemma

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

Let $\gamma: \closedint a b \to S$ be a path in $S$.

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


 * For all $\epsilon \in \openint 0 K$, there exists a normal subdivision $\set {x_0, x_1, \ldots, x_{n - 1}, x_n}$ of the closed interval $\closedint a b$ such that:


 * $\ds \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$


 * and for all $i \in \set {0, 1, \ldots, n - 1}$:


 * $\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$

Here, $\map {B_\epsilon} {\map \gamma {x_i} }$ denotes the open ball of $\map \gamma {x_i}$ with radius $\epsilon$.

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

From Closed Real Interval is Compact in Metric Space, it follows that $\closedint a b$ is compact.

Then Continuous Image of Compact Space is Compact shows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is compact.

From the Heine-Borel Theorem, it follows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ 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:
 * $\map d {\map \gamma {\closedint a b}, \R^m \setminus S} > 0$

Put $K = \map d {\map \gamma {\closedint a b}, \R^m \setminus S}$, the distance between $\map \gamma {\closedint a b}$ and $\R^m \setminus S$.

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

Finding the subdivision
Let $\epsilon \in \openint 0 K$.

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

Then there exists $\delta \in \R_{>0}$ such that:
 * $\forall y, z \in \closedint a b : \size {y - z} < \delta$ implies $\map d {\map \gamma y, \map \gamma z } < \epsilon$

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 $i \in \set {0, 1, \ldots, n}$.

Then $\set {x_0, x_1, \ldots, x_n}$ is a normal subdivision of $\closedint a b$.

It follows that $\map {B_\epsilon} {\map \gamma {x_i} }$ and $\R^m \setminus S$ are disjoint, as either:
 * $\epsilon < \map d {\map \gamma {\closedint a b }, \R^m \setminus S}$

or:
 * $\R \setminus S = \O$

Then:
 * $\ds \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$

Let $x \in \closedint {x_i} {x_{i + 1} }$.

Then:
 * $\size {x - x_i} \le \dfrac{b - a} n < \delta$

As the uniorm continuity condition applies, it follows that:


 * $\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$

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 $\C$.