# REVISITED: What is an open cover?

What exactly is an open cover? Could someone explain it to me like I'm five, and then give a professional answer too?

Here is some literature that may be of interest to any onlooker curious about what an open cover is:

Reference: M.A. Armstrong's Basic Topology

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A collection of open sets that collectively cover another set. – anon Sep 9 '13 at 1:13
What is a finite subcover? – Loie Benedicte Sep 9 '13 at 2:44
A subcover that is finite. – anon Sep 9 '13 at 2:54
Ask a funny question, get a funny answer. A native English-speaking five-year old (perhaps a bit older) will probably be able to understand the concept of (possibly infinitely many) regions in space covering another region, and a finite number of those regions being sufficient to cover it as well, at least with some good visual aids. – anon Sep 9 '13 at 3:05

An open cover of a set $Y$ is a family, (collection), of sets that are open, (a set of open sets), such that $Y$ is a subset of the union of sets in that family.

Of course when I say "open", I mean that these sets are included in some topology $\mathcal T$ on a space $X$, and $Y \subseteq X$.

"Union up all the sets in the family",
"If $Y$ sits inside of that union"
"Then that family is an open cover"

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Is the empty set an open cover of the empty set? Are all sets an open cover of the empty set? – Loie Benedicte Sep 8 '13 at 23:31
@LoieBenedicte The empty set is a subset of all sets so it is certainly a subset of itself. So I would say, "Yes", and "Yes". – Rustyn Sep 8 '13 at 23:33
It's a collection of open sets that covers another one, yes? – Loie Benedicte Sep 8 '13 at 23:36
@LoieBenedicte Yep – Rustyn Sep 8 '13 at 23:37
@LoieBenedicte Actually your open cover of [0,1] would be the set: $\{n\in \mathbb{Z}^{+}: \left(\frac{1}{n}-1,\frac{n+1}{n}+1\right)\}$ – Rustyn Sep 8 '13 at 23:52

Tommy has a bunch of toys, and Sally has some too, so does Kelly, and Johnny. Now Tommy, Sally, Kelly, and Johnny have *their favorite*$^1$ toys, you know because some of them they *don't like*$^2$. If we lay all of everybody's favorite toys on the floor, then we will definitely have *all the toys that Joey has*$^3$.

$1$: The interior of the set.

$2$: The boundary of the set.

$3$: The set which is covered by the union.

An open cover of a set $E$, in the metric space $\chi$, is a collection of sets $\{G_\alpha\}$ whose union "covers" (contains) $E$, and so, for example if you're given the set $E=\{[1,3]\}$ in the metric space $\chi=(\mathbb{R},d)$, where $d$ is the Euclidean metric, then provided the sets $G_1=\{(0,\frac{3}{2})\}$, $G_2=\{(1,\frac{5}{2})\}$, and $G_3=\{(2,4)\}$ we can say that $\bigcup G_\alpha$ covers $E$.

Above we see that $\{G_\alpha\}$, where $\alpha=1,\dots, 5$, is an open cover of $E$. The green lines denote openness. Note that all sets but $G_5$ are necessary to cover $E$, and as such the union of the sets $G_1, G_2, G_3$, and $G_4$ are a subcover of $\{G_\alpha\}$ as they cover $E$.

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Do I have this idea down? – Loie Benedicte Sep 9 '13 at 0:38
I just want to know if I used the language correctly to describe this idea, and if I'm using the notation correctly and what have you. – Loie Benedicte Sep 9 '13 at 1:00
The picture looks good, I didn't read exactly what was written above it. The picture is exemplary... if all of those $G$'s are open. – Rustyn Sep 9 '13 at 1:07