I read the appendix of Support Vector Machines book on the other day and wondered about how we can show that an open set in Rd is not compact. Let me start by giving the definition of compactness and a related theorem. These are from the book.

Definition A set A in a topological space is compact if for every family (Oi)iI of open sets with AiIOi, there exist finitely many indexes i1,,inI with Anj=1Oij. In other words, A is compact if each of its open covers has a finite subcover (from this Wiki page).

Theorem ARd is compact if and only if A is closed and bounded.

Here is what I wondered after reading these two items. Consider an open unit sphere S={xx<1} in Rd. Obviously by the theorem S is not compact because it is not closed. That means there must exist an open cover such that there is no finite subcover. What kind of open cover is that ? That is my question.

I got my answer after discussing with Zoltan. What we need are two things. Firstly, we need to find open covers {Oi}i=1 that can cover S i.e., Si=1Oi. Secondly, show that no finite number of {Oi}i can cover S.

Define Oi=Sri where Sr={xx<r} and (ri)i is a monotonically increasing sequence converging to 1 with ri<1 for all i. No finite number of {Oi}i will cover S because jJOj=Smax({rjjJ}) for any finite index set J and Smax({rjjJ})S. To show Si=1Oi, we need to show that for all pS,pi=1Oi. This is equivalent to: for all pS, there always exists an open set ORp for some R. The latter statement holds since the open cover sphere can grow to any radius <1.