Use the definition of compactness. Take an open covering of . Choose such that . By the definition of convergence of a sequence, outside there are only a finite number of terms of the sequence. Now the conclusion is almost immediate.
Q: Let (a_n) be a sequence of reals so that a = lim(a_n) exists as n→∞. Let S = {a_1, a_2, a_3, · · · } ∪ {a}. Prove that any open cover of S has a finite subcover.
I am trying to prove this without using the Heine-Borel Theorem (Heine?Borel theorem - Wikipedia, the free encyclopedia), which is why it is giving me a little trouble. I'm guessing the goal is basically to arrive at a conclusion showing that S is compact, without using the fact that S is bounded and closed (because that's the Heine-Borel Thm).
My intuition would be to show that S is sequentially compact, so that any sequence in S has a subsequence converging to a limit point in S? I'm not sure where to go from here though... Any hint/help would be appreciated.
Use the definition of compactness. Take an open covering of . Choose such that . By the definition of convergence of a sequence, outside there are only a finite number of terms of the sequence. Now the conclusion is almost immediate.
Just to clarify, so within there would be an infinite # of terms converging to . If so, then S satisfies the definition of compactness that I had on the first post: "any sequence in S has a subsequence converging to a limit point in S" because converges to and any subsequences of converges to . Is this right?
Within , there are infinitely many points of S. What Enrique is saying is that since this one open set covers all but finitely many (say all , for some ), then you can cover remaining points individually.
The "every sequence has a convergent subsequence" (Bolzano-Weierstrass Theorem) is not used in this particular proof; however, it can be used. Since converges to , every subsequence also converges to . Since every sequence has a convergent subsequence, so is compact.