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

  • $\begingroup$ youtube.com/watch?v=zeVA74yivyg $\endgroup$ Sep 8, 2013 at 23:29
  • 1
    $\begingroup$ A collection of open sets that collectively cover another set. $\endgroup$
    – anon
    Sep 9, 2013 at 1:13
  • $\begingroup$ What is a finite subcover? $\endgroup$ Sep 9, 2013 at 2:44
  • 4
    $\begingroup$ A subcover that is finite. $\endgroup$
    – anon
    Sep 9, 2013 at 2:54
  • 3
    $\begingroup$ 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. $\endgroup$
    – anon
    Sep 9, 2013 at 3:05

2 Answers 2


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"

  • $\begingroup$ Is the empty set an open cover of the empty set? Are all sets an open cover of the empty set? $\endgroup$ Sep 8, 2013 at 23:31
  • $\begingroup$ @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". $\endgroup$
    – Rustyn
    Sep 8, 2013 at 23:33
  • $\begingroup$ It's a collection of open sets that covers another one, yes? $\endgroup$ Sep 8, 2013 at 23:36
  • $\begingroup$ @LoieBenedicte Yep $\endgroup$
    – Rustyn
    Sep 8, 2013 at 23:37
  • 1
    $\begingroup$ @LoieBenedicte Actually your open cover of [0,1] would be the set: $\{n\in \mathbb{Z}^{+}: \left(\frac{1}{n}-1,\frac{n+1}{n}+1\right)\}$ $\endgroup$
    – Rustyn
    Sep 8, 2013 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 $E$.

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 $E$.

  • $\begingroup$ Do I have this idea down? $\endgroup$ Sep 9, 2013 at 0:38
  • $\begingroup$ 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. $\endgroup$ Sep 9, 2013 at 1:00
  • 1
    $\begingroup$ 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. $\endgroup$
    – Rustyn
    Sep 9, 2013 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.