The Lindelöf Covering Theorem in Euclidean Space

# The Lindelöf Covering Theorem in Euclidean Space

Recall from the Coverings of a Subset in Euclidean Space page that if $S \subseteq \mathbb{R}^n$ then a covering of $S$ is a collection of subsets from $\mathbb{R}^n$, $\mathcal F$, such that:

(1)\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal F} A \end{align}

If every set in $\mathcal F$ is open then $\mathcal F$ is called an open covering of $S$ and if every set in $\mathcal F$ is closed then $\mathcal F$ is called a closed covering of $S$.

Lemma 1: Let $S \subseteq \mathbb{R}^n$ be an open set and let $\mathbf{x} \in \mathbb{R}^n$. Let $\mathcal A = \{ A_1, A_2, ... \}$ be the countable collection of open balls of the form $B(\mathbf{y}, q)$ where $\mathbf{y} = (y_1, y_2, ..., y_n)$ and $q > 0$ are such that $y_1, y_2, ..., y_n, q \in \mathbb{Q}$. Then there exists at least one $A_i \in \mathcal A$ such that $\mathbf{x} \in A_i \subseteq S$. |

**Proof:**Let $S \subseteq \mathbb{R}^n$ be an open set. Then $S = \mathrm{int} (S)$. If $\mathbf{x} \in \mathbb{R}^n$ is contained in $S$, then this means that for some positive real number $r > 0$ there exists a ball centered at $\mathbf{x}$ with radius $r > 0$ such that:

\begin{align} \quad B(\mathbf{x}, r) \subseteq S \end{align}

- Let $\mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$, and for each $k \in \{ 1, 2, ..., n \}$ let $y_k \in \mathbb{Q}$ be chosen so that:

\begin{align} \quad \mid y_k - x_k \mid = \mid x_k - y_k \mid < \frac{r}{4n} \end{align}

- We now that a rational $y_k$ can always be chosen such that $\mid x_k \mid - \frac{r}{4n} < y_k < \mid x_k \mid + \frac{r}{4n}$ by The Density of Real Numbers Theorem page.

- If the coordinates of $\mathbf{y} \in S$ are chosen this way, then the distance between $\mathbf{x}$ and $\mathbf{y}$ is bounded by:

\begin{align} \quad \| \mathbf{x} - \mathbf{y} \| = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + ... + (x_n - y_n)^2} \\ \quad \| \mathbf{x} - \mathbf{y} \| \leq \mid x_1 - y_1 \mid + \mid x_2 - y_2 \mid + ... + \mid x_n - y_n \mid \\ \quad \| \mathbf{x} - \mathbf{y} \| \leq \sum_{k=1}^{n} \mid x_k - y_k \mid \\ \quad \| \mathbf{x} - \mathbf{y} \| < n \cdot \frac{r}{4n} = \frac{r}{4} \end{align}

- Now choose $q \in \mathbb{Q}$ such that:

\begin{align} \quad \frac{r}{4} < q < \frac{r}{2} \end{align}

- Such a rational number $q$ exists by the density of the real numbers. Since $\frac{r}{4} < q < \frac{r}{2} < r$ and $\| \mathbf{x} - \mathbf{y} \| < \frac{r}{4}$ we must have that $\mathbf{x} \in B(\mathbf{y}, q)$. Furthermore the distance from $\mathbf{x}$ to $\mathbf{y}$ is less than $\frac{r}{4}$, and the distance from any point from $\mathbf{y}$ in $B(\mathbf{y}, q)$ is less than $\frac{r}{2}$, so:

\begin{align} \quad \mathbf{x} \in B(\mathbf{y}, q) \subseteq B(\mathbf{x}, r) \end{align}

- Since $y_1, y_2, ..., y_n, q \in \mathbb{Q}$ we have that $B(\mathbf{y}, q) = A_i \in \mathcal A$ for some $i \in \{1, 2, ... \}$ which completes our proof $\blacksquare$

Theorem 1 (The Lindelöf Covering Theorem): If $S \subseteq \mathbb{R}^n$ and $\mathcal F$ is an open covering of $S$ then there exists a countable subcovering of $\mathcal F$ for $S$. |

**Proof:**Let $\mathcal A = \{ A_1, A_2, ... \}$ be the countable collection of open balls of the form $B(\mathbf{y}, q)$ where $\mathbf{y} = (y_1, y_2, ..., y_n)$ and $q > 0$ are such that $y_1, y_2, ..., y_n, q \in \mathbb{Q}$ and let $\mathbf{x} \in S$. Since $\mathcal F$ is an open covering of $S$ there exists an open set $F \in \mathcal F$ such that $\mathbf{x} \in F$.

- By Lemma 1, since $F$ is an open set, there exists an $A_i \in \mathcal A$ such that:

\begin{align} \quad \mathbf{x} \in A_i \subseteq F \end{align}

- For each $S$ let $A_{f(\mathbf{x})}$ be the smallest indexed member of the collection $\mathcal A$ such that $\mathbf{x} \in A_{f(\mathbf{x})} \subseteq F$. Then for all $\mathbf{x} \in S$, the collection $\{ A_{f(\mathbf{x})} \}_{\mathbf{x} \in S}$ is a countable collection of open sets such that:

\begin{align} \quad S \subseteq \bigcup_{\mathbf{x} \in S} A_{f(\mathbf{x})} \end{align}

- So $\displaystyle{\bigcup_{\mathbf{x} \in S} A_{f(\mathbf{x})}}$ is a countable open covering of $S$. Correlate each $A_{f(\mathbf{x})}$ to each $F \in \mathcal F$ to get a countable subcovering of $\mathcal F$ for $S$. $\blacksquare$