REVISITED: What is an open cover?

Question

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:

enter image description here

Reference: M.A. Armstrong's Basic Topology

A collection of open sets that collectively cover another set. — anon, Sep 09 '13 at 01:13
@anon, OMG, really? I mean what does finite mean explicitly? Are you an American, anon? — Loie Benedicte, Sep 09 '13 at 02:59
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 09 '13 at 03:05

Rustyn · Answer 1 · 2013-09-08T23:45:32.410

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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"

edited Sep 08 '13 at 23:45

answered Sep 08 '13 at 23:29

Rustyn

8,407

Is the empty set an open cover of the empty set? Are all sets an open cover of the empty set? – Loie Benedicte Sep 08 '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 08 '13 at 23:33
It's a collection of open sets that covers another one, yes? – Loie Benedicte Sep 08 '13 at 23:36
@LoieBenedicte Yep – Rustyn Sep 08 '13 at 23:37
Hey Rus,I don't think you have explained it to a five year old yet (Paul here, I should probably change my name). – Sep 08 '13 at 23:38
So an open cover of $[0,1]$ could be $\bigcup_{n=1}^{\infty}\pmatrix{\frac{1}{n}-1,\frac{n+1}{n}+1}$? – Loie Benedicte Sep 08 '13 at 23:40
@Rustyn Sure want to create a room, or google chat or something – Sep 08 '13 at 23:42
@LoieBenedicte ${(-1,2)}$ is also an open cover. – Rustyn Sep 08 '13 at 23:48
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@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 08 '13 at 23:52
@PaulPlummer I invited you to chat thru broncomail – Rustyn Sep 08 '13 at 23:57
I see, so the set ${n\in\mathbb{N}:\pmatrix{\frac{1}{n},1-\frac{1}{n}}}$ is not an open cover of $[\frac{1}{2},1]$, right? – Loie Benedicte Sep 09 '13 at 00:04
@LoieBenedicte Yep it is "NOT" an open cover, because $1$ is not covered. – Rustyn Sep 09 '13 at 00:05
Is there an open cover of $\mathbb{R}$? – Loie Benedicte Sep 09 '13 at 00:07
@LoieBenedicte Of course, set the open cover $\mathcal O$ = ${\mathbb{R}}$ – Rustyn Sep 09 '13 at 00:08
@PaulPlummer -- were you implying that you are 5 years old? Or do you have a five year old kid who still doesn't understand this? – Betty Mock Sep 09 '13 at 01:03
@BettyMock The question says to explain to the op like Loie was five years old – Sep 09 '13 at 01:27
@BettyMock I was asking in the spirit of the following site: http://www.reddit.com/r/explainlikeimfive/ – Loie Benedicte Sep 09 '13 at 01:36
This comment thread is going to infuriate the moderators of this website. I think this should be the last comment. – Rustyn Sep 09 '13 at 01:38
can Y be exactly that union? and does any subset has an open cover? – user38448-po Dec 24 '13 at 13:07

Loie Benedicte · Answer 2 · 2013-09-09T01:34:38.433

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 $E$.

enter image description here

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 $E$.

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 09 '13 at 01:00
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 09 '13 at 01:07

REVISITED: What is an open cover?

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