Let (x_n) be a sequence, let $L$ ∈ R, and for each ϵ>0,
{k ∈ N : x_k ∈ B($L$; ϵ)}
Suppose S is not a compact subset of R. There is some ϵ_L > 0, such that
{k ∈ N : x_k ∈ B($L$; ϵ_L)} is finite.
Question: Let O:={B($L$,ϵ_L): $L$ ∈S}. Explain why $O$ is an open cover for S that has no finite subcover.
Note: I understand for something like [0,1], and finding some open covers. Also that it has finitely many subcovers. But I am confused with the open cover of something more general.
I see how it is possible, but I dont know how to show it.
I know that the intersection of infinitely many open sets may fail to be open.
Best Answer
Think of the pigeonhole principle. We have an infinite collection of numbers $x_n$, and infinitely many boxes to put them in: the collection of open sets in $O$. If a find subcover existed, we would be stuffing our infinite collection of numbers into finitely many open sets. One of these open sets in $O$ would have to have infinitely many elements of $x_n$. This gives a contradiction.