(Definition 3.3.6) Let $A \subseteq \mathbf{R}$. An open cover for $A$ is a (possibly infinite) collection of open sets $\{O_{\lambda}: \lambda \in \Lambda\}$ whose union contains the set $A$; that is, $A \subseteq_{\lambda \in \Lambda} O_{\lambda}$. Given an open cover for $A$, a finite subcover is a finite sub collection of open sets from the original open cover whose union still manages to completely contain $A$.

Example

An open cover of $(0,4)$ is the following

$$ \begin{align*} \left\{\left(\frac{1}{k}, 4 - \frac{1}{k}\right)\right\}^{\infty}_{k=1} = \left\{(1,3), (\frac{1}{2},4-\frac{1}{2}),...\right\}. \end{align*} $$

Each set above is open. Note here that we have to have an infinite collection of these sets. Any finite subset will not contain all of $(0,4)$.

An open cover of $[0,4]$ is the following

$$ \begin{align*} \left\{\left(\frac{1}{k}, 4 - \frac{1}{k}\right)\right\}^{\infty}_{k=1} \bigcup \quad \{(-0.2,0.2),(3.8,4.2)\}. \end{align*} $$

Note here that we had to add two extra sets to cover both end points. This illustrates the difference between what we need to cover an open interval versus a a closed interval.

What about a finite subcover of the following open cover of $(0,4)$

$$ \begin{align*} \left\{\left(\frac{1}{k}, 4 - \frac{1}{k}\right)\right\}^{\infty}_{k=1} = \left\{(1,3), (\frac{1}{2},4-\frac{1}{2}),...\right\}. \end{align*} $$

Is it possible to get a finite cover that covers $(0,4)$? No, in this case no matter what collection we choose, we’ll always be missing an element from the interval $(0,4)$.

Other Definitions and Properties

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