# Does a cover have to be infinite

I have encountered this question: find a cover for $[1,2]$ in $\mathbb{R}$ that does not have a finite subcover.

Since $[1,2]$ is closed and bounded then it's compact. It then implies that a cover that does not have a finite subcover must not be open. I was thinking about $[1,1.5] \cup [1.5,2]$. However, it is not an infinite union of sets. Can it still be a cover?

• If covers had to be infinite, there would be no such thing as a finite subcover; subcovers have to be covers. – user2357112 Dec 5 '17 at 17:50

A cover does not have to be infinite, and the example you gave is in fact a finite cover. A cover of a set $S$ is simply a collection $\mathcal{F}$ of sets such that for any $x \in S$, there is at least one $A \in \mathcal{F}$ with $x \in A$. There can be any number of sets in $\mathcal{F}$, and they don't even need to be disjoint. But there are a few issues with the example you gave, and how your finite cover relates to it:

1. Since the cover you gave is finite, it does have a finite subcover: the cover itself.
2. What you want to do is find a particular infinite cover $\mathcal{F}$ such that there does not exist a finite subset of $\mathcal{F}$ that is also a cover.
3. In most cases (i.e. for showing compactness) you want $\mathcal{F}$ to be an open cover, that is, you want the sets in $\mathcal{F}$ to all be open.
• I decided to go with your answer since you answered my questions. – user1691278 Dec 5 '17 at 5:15
• This is a good, detailed answer but I am sad :( – Levent Dec 5 '17 at 5:16
• @Levent Don't be sad. I think my answer and Morgan Rodgers' answer perfectly complement each other. The OP probably has benefitted from both, which is the thing that is important here on MSE. – user491874 Dec 5 '17 at 9:34

Just cover it with one-element sets: $\{\{x\}, x\in[1,2]\}$. This is uncountable infinite cover and you cannot omit even a single one of them!

• I will be sad if this answer won't get accepted. – Levent Dec 5 '17 at 5:02
• @Levent This answer being accepted actually makes me confused what the OP was asking. shrugs (but it's a good clever example) – Morgan Rodgers Dec 5 '17 at 5:07
• @MorganRodgers Actually, you made a good point. This answer does have a finite subcover - the cover itself. – user1691278 Dec 5 '17 at 5:07
• @MorganRodgers Actually, the cover is infinite though... Let me think more about it. – user1691278 Dec 5 '17 at 5:09
• @user1691278 No this example does not have a finite subcover; the cover itself is not a finite subcover because it is not finite. But I am not sure if your original question was asking a: for an example of a cover of $[1,2]$ without a finite subcover or b: if it is acceptable for a cover of a set to be finite or c: whether or not the finite set you gave was a solution to the question posed in the first line – Morgan Rodgers Dec 5 '17 at 5:09

It's actually very simple. Just take $[1,1.1],[1.9,2]$ to be in the cover. Now what is left over is $(1.1,1.9)$, which is not closed, so not compact. Hence, there is an open cover not having a finite subcover. For example, $(1.1 + \frac 1n, 1.9-\frac 1n)_{n \in \mathbb N}$ would do. Note that these are nested, so the union of the sets of any finite subcover, is just the largest set of that cover, but none of the sets are equal to $(1.1,1.9)$, so this qualifies.

So, $\mathcal U = \{[1,1.1],[1.9,2],\left(1.1 + \frac 1n,1.9-\frac 1n\right)_{n \in \mathbb N}\}$ works as a infinite cover having no finite subcover. Note it does contain non-open sets, as you remarked earlier.

EDIT : This construction is of a "countable" cover not having a finite subcover. For a nice trivial example, the other answer does well.

• Your example is brilliant. Does my example work though, if a cover does not have to be an infinite union of sets? – user1691278 Dec 5 '17 at 4:58
• Yes, your example works. – астон вілла олоф мэллбэрг Dec 5 '17 at 5:00
• @астонвіллаолофмэллбэрг His example does not work, it contains a finite subcover. – Morgan Rodgers Dec 5 '17 at 5:00
• @MorganRodgers But $\mathcal U = \{[1,1.5],[1.5,2]\}$ does not have a strictly smaller subcover, right? – астон вілла олоф мэллбэрг Dec 5 '17 at 5:02
• @астонвіллаолофмэллбэрг A subcover does not need to be strictly contained in the cover (a set is a subset of itself, for example) – Morgan Rodgers Dec 5 '17 at 5:03