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Definition of an open cover is defined as follows in my book (Advanced Calculus by Leonard F. Richardson):

An open cover of a set $S \subseteq \mathbb{R}$ is a collection $$\mathcal{O} = \{O_\alpha|\alpha \in A \}$$ of (perhaps infinitely many) open sets $O_\alpha$, where $\alpha$ ranges over some index set $A$, such that $S \subseteq \cup_{\alpha \in A}O_\alpha$.

I am not sure if the open cover is an union of open sets, as the definition calls it a collection of subsets. Then, it's possible the open cover could be a collection of disjoint open sets like below?

My understanding of open cover

But for any element $x$ in an open cover, it will be one of $O_1, O_2, O_3$. Since $O_1, O_2, O_3$ are open sets, we can find an open interval that is contains $x$ in one of $O_1, O_2, O_3$. So, open cover should follow the definition of an open set. But I remember from complex analysis course, that open sets cannot be a union of disjoint open sets like my drawing.

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    The union of the sets in the cover is an open set. The open cover as a collection of sets isn't an open set, unless there's only one open set in the collection. – 2017-01-23
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    @user4894 Technically speaking, even if there's only one open set in the collection, the cover is not an open set. It is a set containing an open set. – 2017-01-23
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    @ZacharySelk But doesn't open cover satisfy the definition of an open set? If I pick any element in the open cover, the element is contained in one of the open sets that are in the open cover. Thus, there is an open set that contains that element and therefore the open cover has an open set that contains the element. – 2017-01-23
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    This is a set theory thing not anything to do with topology. An open cover is a collection of open sets. – 2017-01-23
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    Is it because I can only take open set as an element of the open cover? I just realized this – 2017-01-23
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    Does this mean the set $S \subseteq \mathbb{R}$ in the definition is also a set of open sets? – 2017-01-23
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    @user3000482 No of course not, $S$ is a single subset of $\mathbb{R}$ (ie a set of real numbers, not a set of sets) – 2017-01-23
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    @ZacharySelk Right you are. Thanks for the correction. – 2017-01-23

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An open cover is a COLLECTION of open sets, so it is not an open set.

An open set of $\Bbb R$ is a set $S\subset \Bbb R$ that satisfies some conditions. Note that $\mathcal O \not \subset \Bbb R$ so $\mathcal O$ is not an open set.

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    Does that mean $S \subseteq \mathbb{R}$ is not an open set, but also a set of open sets? – 2017-01-23
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    @user3000482 No $S$ is a set of numbers – 2017-01-23
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    Is $S$ going to be an open set? Or it could be an open set as well as it could not be an open set? – 2017-01-23
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    @user3000482 I think there's a miscommunication. I'm saying a set $S\subset \Bbb R$ is open if it satisfies some conditions. Your $\mathcal O$ is not a subset of $\Bbb R$. – 2017-01-23
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    If every element of $S$ is in the union of open sets that are in the open cover that contains those open sets, then $S$ will be an open set right? – 2017-01-23
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    @user3000482 No this is not true because the union of open sets might be strictly larger than $S$ itself. For example you might have $S = [0,1]$, and the union of the open sets might be the open set $(-.5,1.5)$ containing $S$. – 2017-01-23