Open Cover and Compactness I'm having trouble when thinking about compactness.
We knew that compactness of $A\subseteq\mathbb{R}$, can be defined as:
Every open cover of $A$ has a finite subcover.
My question is:
An open cover is a collection of open sets.
So, when we are talking about the finite subcover, can I have a number of countably infinite  intervals, here?
Because a finite subcover means a finite open sets, and every open set in $\mathbb{R}$ is characterized as disjoint union of a countably open intervals.
Any help will be appreciate.
 A: Here we need to distinguish two quantities that we are counting:

*

*The number of open sets in the open cover

*The number of intervals that each open set can be decomposed into

Now certainly as you have suggested, every open set in $\mathbb{R}$ can be decomposed into countably many open intervals. This is related to the second quantity in the above.
However, the word finite in the definition of compact refers to the first quantity in the above. In other words, by finite subcover, we mean that the open sets are chosen from the original open cover and the number of such sets are finite.
So whether the open sets can be further decomposed into open intervals is irrelevant in the definition of compactness.
Edit: In response to the OP's follow up question, I would like to use a rather trivial example to emphasize that it is possible for some open set in a particular finite open cover to be a disjoint union of infinitely many open intervals.
Consider the set $ [0,1]$ in $ \mathbb{R}$.
This set is compact.
Now consider this specific open cover
$$ \mathcal{O} = \lbrace (-1,2), [0,1]-\mathcal{C} \rbrace, $$
where $\mathcal{C}$ is the Cantor set.
There are two open sets in $ \mathcal{O} $, so $ \mathcal{O} $ is a finite open cover of $ [0.1] $.
But the set $ [0,1] -\mathcal{C} $ is a disjoint union of countably many open intervals.
In particular, the number of open intervals is not finite.
In addition, if we write out
$$ [0,1] - \mathcal{C} = \bigcup_i I_i $$
explicitly, where each of the $I_i$ is an open interval. Then actually each of the $I_i$ is not an element of the open cover $\mathcal{O}$.
A: It might help to write out the definition of compactness in a form commonly used in applications: Let $\Lambda$ be an arbitrary set. If $\{O_\lambda \mid \lambda \in \Lambda\}$ is a collection of open sets indexed by $\Lambda$ such that $A \subset \bigcup_{\lambda \in \Lambda}O_\lambda$, then there exists a finite subset $\{\lambda_1, \dots, \lambda_n\} \subset \Lambda$ ($n \geq 0$) such that $A \subset \bigcup_{i = 1}^{n}O_{\lambda_i}$.
