# Coverings of a Set in a Metric Space

Definition: Let $(M, d)$ be a metric space and let $S \subseteq M$. A Covering of $S$ is a collection of sets $\mathcal F$ from $M$ such that $\displaystyle{S \subseteq \bigcup_{A \in \mathcal F} A}$ and $\displaystyle{\bigcup_{A \in \mathcal F} A}$ is called a Cover of $S$. A Subcovering of $S$ is a subcollection of sets $\mathcal S \subseteq \mathcal F$ such that $\displaystyle{S \subseteq \bigcup_{A \in \mathcal S} A}$. |

Definition: An Open Covering of $S$ is a collection of open sets $\mathcal F$ from $M$ such that $\displaystyle{S \subseteq \bigcup_{A \in \mathcal F} A}$. An Open Subcovering of $S$ is a subcollection of open sets $\mathcal S \subseteq \mathcal F$ such that $\displaystyle{S \subseteq \bigcup_{A \in \mathcal S} A}$. |

For example, consider the metric space $(\mathbb{R}, d)$ where $d$ is the Euclidean metric on $\mathbb{R}$. Consider the subset $S = (0, 1) \subseteq \mathbb{R}$. One such covering of this interval is the collection of sets:

(1)Each of the sets in $\mathcal F$ is open. To prove that $\mathcal F$ is an open covering of $(0, 1)$ we must show that:

(2)Let $x \in (0, 1)$. There exists a natural number $n(X) \in \mathbb{N}$ such that $0 < \frac{1}{x} < n(x)$ and so:

(3)So $\displaystyle{x \in \left ( \frac{1}{n(x)}, 1 \right) \subseteq \bigcup_{n=1}^{\infty} \left ( \frac{1}{n}, n \right )}$. Therefore $\displaystyle{x \in \bigcup_{n=1}^{\infty} \left ( \frac{1}{n}, n \right )}$ so $\displaystyle{S \subseteq \bigcup_{n=1}^{\infty} \left ( \frac{1}{n}, n \right )}$ and $\displaystyle{\left \{ \bigcup_{n=1}^{\infty} \left ( \frac{1}{n}, n \right ) : n \in \mathbb{N} \right \}}$ is an open covering of $S = (0, 1)$.

Definition: If $(M, d)$ is a metric space and $S \subseteq M$ then a covering $\mathcal F$ of $S$ is said to be Finite if $\mathcal F$ is a finite collection of sets. $\mathcal F$ is said to be Countable if $\mathcal F$ is countable collection of sets. $\mathcal F$ is said to be Uncountable if $\mathcal F$ is an uncountable collection of sets. |

In the example above, we see $\displaystyle{\mathcal F = \left \{ \bigcup_{n=1}^{\infty} \left ( \frac{1}{n}, n \right ) : n \in \mathbb{N} \right \}}$ is a infinite countable open covering on $S$ since we can define a bijection $f : \mathbb{N} \to \mathcal F$ for all $n \in \mathbb{N}$ by:

(4)