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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?

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    $\begingroup$ If covers had to be infinite, there would be no such thing as a finite subcover; subcovers have to be covers. $\endgroup$ Dec 5, 2017 at 17:50

3 Answers 3

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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!

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  • $\begingroup$ I will be sad if this answer won't get accepted. $\endgroup$
    – Levent
    Dec 5, 2017 at 5:02
  • $\begingroup$ @MorganRodgers Actually, you made a good point. This answer does have a finite subcover - the cover itself. $\endgroup$ Dec 5, 2017 at 5:07
  • $\begingroup$ @MorganRodgers Actually, the cover is infinite though... Let me think more about it. $\endgroup$ Dec 5, 2017 at 5:09
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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.
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    $\begingroup$ I decided to go with your answer since you answered my questions. $\endgroup$ Dec 5, 2017 at 5:15
  • $\begingroup$ This is a good, detailed answer but I am sad :( $\endgroup$
    – Levent
    Dec 5, 2017 at 5:16
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    $\begingroup$ @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. $\endgroup$
    – user491874
    Dec 5, 2017 at 9:34
  • $\begingroup$ Since we're talking about an "open cover" and [1,2] is compact, shouldn't there always be a finite subcover then? Can you please explain point 2? $\endgroup$
    – ali
    Nov 3, 2021 at 9:19
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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.

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  • $\begingroup$ Your example is brilliant. Does my example work though, if a cover does not have to be an infinite union of sets? $\endgroup$ Dec 5, 2017 at 4:58
  • $\begingroup$ Yes, your example works. $\endgroup$ Dec 5, 2017 at 5:00
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    $\begingroup$ @MorganRodgers But $\mathcal U = \{[1,1.5],[1.5,2]\}$ does not have a strictly smaller subcover, right? $\endgroup$ Dec 5, 2017 at 5:02
  • $\begingroup$ @MorganRodgers Yes, you are correct there. I should have mentioned it earlier. Thank you. $\endgroup$ Dec 5, 2017 at 5:04

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