Compactness of A: Proving it is a Subset of R

[Buri]

Let A = {0,1,1/2,1/3,...,1/n,...}. Prove that A is a compact subset of R.

Proof:

Let {U_i} be an open cover for A. Therefore, there must exist a U_0 such that 0 is in U_0. Now since, U_0 is open and 1/n converges to 0, there must be infinite number of points of A in U_0. Now by the well ordering principle U_0 must contain a largest element (Note that {k, k+1,...} has a least element implies {1/k,1/(k+1),...} has a largest element <== this is probably what I'm not 100% sure of about the largest element)...So the points 1/(m-1), ..., 1 will only need AT MOST m-1 more open sets to cover. Therefore, a finite sub cover of {U_i} exists.

Is this proof correct?

 

[radou]

I don' quite see why you invoked the well ordering principle here. If you take any open cover for A, 0 must be in at least one element of this cover, let's say in some U. Since 1/n converges to 0, U must contain all but a finite number of elements of this sequence. For every one of these elements, there must exist at least one set containing that element, so if you take the union of U and this finite collection of sets, you have found an open subcover of your original cover which is finite.

 

[Buri]

Hmm I see what you mean. Well my problem was that how do I really know that from some point on those elements are no longer in U. But I guess the convergence of 1/n basically covers it. Hopefully I don't get penalized for it lol Thanks a lot for your help.

 

[radou]

Hmm I see what you mean. Well my problem was that how do I really know that from some point on those elements are no longer in U. But I guess the convergence of 1/n basically covers it. Hopefully I don't get penalized for it lol Thanks a lot for your help.

You know this by the definition of convergence of a sequence. There exists some positive integer N such that, for all n >= N, the members of the sequence are in U. Hence, at most n - 1 members of your sequence are not in U, which is definitely a finite number. (At most because the set U which contains 0 can be a union of a number of sets, so there can be some other elements which are in U, but for some n < N, they "jump out" of U again.)

 

[Buri]

Yeah, I messed up :( Oh well, its just one problem set lol Thanks for the help!

 

[HallsofIvy]

Hmm I see what you mean. Well my problem was that how do I really know that from some point on those elements are no longer in U.

You don't! It might well happen that all points of the set are in U. But that's not a problem- that just means that U itself would be a finite subcover. The important point is that there are at most a finite number of points that are not in U.

 

[Buri]

Thanks for helping me once again :)