Cover¶

Definition¶

Consider a (topological) space $X\text{.}$ Let $E$ be a subset of $X\text{,}$ and let $\mathcal{C}=\{G_i\colon i\in I\}$ be a collection of subsets $G_i$ of $X$ indexed by an arbitrary index set $I\text{.}$ The collection $\mathcal{C}$ is called a cover of $E$ if $\displaystyle E \subseteq \bigcup_{i\in I} G_i\text{.}$

• An open cover is a cover where each "patch" $G_i$ is open.
• A finite cover consists of a collection $\{G_i\colon i\in I\}$ where $I$ is finite. A cover is called a countable cover when $I$ is a countable set, e.g., $\mathbb{N}\text{.}$
• A subcover is a cover that is a subset of another cover.

The Union of Sets collects all points lying in at least one patch $G_i$; the inclusion requires every point of $E$ to occur in that union.

Interpretation¶

A cover is just a collection of sets that together contain the set you care about. Think of it as laying patches over the set so that every point is under at least one patch; the patches may overlap and may stick out beyond the set.

Five open intervals $G_1,\dots,G_5$ lie over $E=[0,1]$$[0,1]$, and every point of the segment is under at least one of them. The outer two reach beyond $0$ and $1$, as they must: an open interval containing $0$ also contains points to the left of $0$. Since each $G_i$ is open and there are finitely many, this is a finite open cover.

The index set $I$ only labels the patches and may be infinite. A subcover keeps some of the patches, but it must still cover $E$: patches can be dropped only as long as no point of $E$ is left uncovered.

Covers matter because of compactness: a set is compact if every open cover has a finite subcover. The first Example shows how this fails for $(0,1)$. The intervals $(\tfrac{1}{n},1-\tfrac{1}{n})$ increase with $n$, so finitely many of them cover only the largest one, $(\tfrac{1}{N},1-\tfrac{1}{N})$$\frac{1}{N})$, and no finite subcover exists. For $[0,1]$, by contrast, every open cover can be reduced to finitely many patches, as the Heine-Borel theorem guarantees.

A common slip is to read "every open cover has a finite subcover" as "some finite open cover exists". The second is always true, because the space $X$ itself covers $E$, as the Key Results note.

Key Results¶

• Any set $E\subseteq X$ is covered by the space $X$. So covers always exist.

Examples¶

• The set $(0,1)$ in $\mathbb{R}$ can be covered by the collection of open sets $\{(1/n,1-1/n)\colon n\geq 3\}$$3\}$. This is an infinite open cover.
• The set $\{1,2,3\}$ is covered by the three singletons $\{1\}$, $\{2\}$, $\{3\}$. That is a finite cover.
• The set of rational numbers $\mathbb{Q}$ in $\mathbb{R}$ can be covered by open intervals centred at each rational with radius $1$. That's a countable cover because $\mathbb{Q}$ is countable.

Prerequisites¶

• Union of Sets
• Open and Closed Sets in a Metric Space
• Countable and Uncountable Sets

Related Concepts¶

• Compact Set
• Heine-Borel Theorem
• Sequentially Compact Set

Sources¶

• Math III: Analysis · Lecture 9

Symbols: element of, natural numbers, rational numbers, real numbers, subset

Last updated: 20 October 2026 03:02