This is concerning the topological definition of a compact set. I’m trying to wrap my head around the def but am presently failing. For specifics, see the wiki article.
Here is my problem: Take the closed interval [0,1] in R which I know a priori to be compact. I can’t figure out how to make this interval from a finite subset of open sets that cover it. Aren’t the interval endpoints a big problem? Because by the “standard” definition of open sets, any open set that contains 0 or 1 will also contain “near enough” points outside the interval. So how do you create a finite cover that also contains the endpoints of the interval? What am I missing here?
Its been a while since I took courses in this, but I think that there are two ways to answer your question. If we are viewing [0,1] as a compact subset of the reals, then you can take an open cover that includes elements outside the set [0,1]. For examples (-1,0.5),(0.25,2) would be an open cover.
If you are viewing [0,1] as its own space (no numbers outside it). Then the definition of open set changes, so that [0,0.5) is an open set. Since there are no numbers less than 0, it contains all numbers that are within a distance delta<0.5 of the number 0, and it is open in the usual sense for all other numbers.
Looking at it as a subset of R rather than it’s own topological space.
But if the subcover can contain more points than the set in question, why isn’t every set compact since it will be covered by the single open set, R itself?
The idea is not that there is some finite open cover. The idea is that every open cover, finite or not, contains a finite subcover. That is, no matter how you pick a collection of open sets whose union contains [0, 1] (and possibly other points as well), it will turn out that in fact some finite subcollection accomplishes the deed just as well.
Yes. Here is an example of an open set, and a cover that has no finite sub cover.
the open set set (0,1) is covered by the sequence of open sets (1/n,1) for n= 2,3,4,… since for any value on (0,1) I can find a m such that 1/m is less than that value and so that value will be covered by (1/m,1). But for any finite subset of these there will be some N that such that (1/N,1) is the last member of the subset, but then the value of (1/2N) wouldn’t be covered. Thus this is an open cover with no finite sub-cover. This doesn’t work for [0,1] because 0 is not covered by (1/n,1) for any n.
I think this is your problem: You’re talking about a topological space, like [0, 1], being compact. So when you’re talking about open sets, you’re talking about subsets of that space being open within that space. There are no “points outside the interval.” For example, if the space you’re considering is [0, 1], something like [0, 0.5) would be an open subset of that space.
Thanks for the replies; I do understand now. I think part of the reason I was looking for covers that are equal to the set in question instead of simply being a super-set of it was because the wikipedia article has the definition phrased that way. But on closer inspection, I saw that the article was discussing compactness of a space, and not a set.
The question of being a set in a topological space or being a space all to itself is definitely tricky but I think was able to keep more or less sane about it
