# Proving compactness without using Heine-Borel Thm

• Nov 11th 2009, 03:51 PM
DPMachine
Proving compactness without using Heine-Borel Thm
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.
• Nov 11th 2009, 04:09 PM
Enrique2
Use the definition of compactness. Take $(U_i)_i$ an open covering of $S$. Choose $i_0$ such that $a\in U_{i_0}$. By the definition of convergence of a sequence, outside $U_{i_0}$ there are only a finite number of terms of the sequence. Now the conclusion is almost immediate.
• Nov 11th 2009, 04:56 PM
DPMachine
Quote:

Originally Posted by Enrique2
Use the definition of compactness. Take $(U_i)_i$ an open covering of $S$. Choose $i_0$ such that $a\in U_{i_0}$. By the definition of convergence of a sequence, outside $U_{i_0}$ there are only a finite number of terms of the sequence. Now the conclusion is almost immediate.

Just to clarify, so within $U_{i_0}$ there would be an infinite # of $a_{n}$ terms converging to $a$. 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 $a_{n}$ converges to $a$ and any subsequences of $a_{n}$ converges to $a$. Is this right?
• Nov 11th 2009, 06:55 PM
redsoxfan325
Quote:

Originally Posted by DPMachine
Just to clarify, so within $U_{i_0}$ there would be an infinite # of $a_{n}$ terms converging to $a$. 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 $a_{n}$ converges to $a$ and any subsequences of $a_{n}$ converges to $a$. Is this right?

Within $U_{i_0}$, there are infinitely many points of S. What Enrique is saying is that since this one open set covers all but finitely many $x_n\in S$ (say all $n>N$, for some $N$), then you can cover remaining points $a_1,...,a_N$ individually.

The "every sequence has a convergent subsequence" (Bolzano-Weierstrass Theorem) is not used in this particular proof; however, it can be used. Since $\{a_n\}$ converges to $a$, every subsequence also converges to $a$. Since $S=\{a_n\}\cup a$ every sequence has a convergent subsequence, so $S$ is compact.