Let be an open cover for . There exists such that . Since the sequence converges to contains all but finitely many elements of , and the rest can obviously be covered by a finite subcover.
Let .Prove that is compact directly from the definition(without using Heine-Borel theorem).
The definition of compactness I'm using is: A set is said to be compact if every open cover of has a finite subcover.
But the problem I'm having is, how do I show that universal statement is true. I can think of open covers which have finite subcovers, but how can I show that it will be the case for any and all open covers without referencing the Heine-Borel theorem in any way.
You can show that the subcovers are finite by finding an arbitrary one.
You can argue:
At the point 0 for some large N, we have that 1/N<L, L some element of K. ie. 0 is a limit point of K. Pick a subcover that covers 0 and 1/N. Note that this covers infinitely many points between 0 and 1/N. That means that you only have finitely many points remaining: ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N-1). Pick one subcover to cover each of these points. Now you have a finite set of subcovers.
Thus, the open set that contains has some radius , and therefore all elements with will be contained in this set. But the set of all natural numbers larger than some fixed constant is clearly infinite.
The remaining points that need to be covered are all the natural numbers less than , which is clearly a finite set and will be covered by at most open sets.
Then what you said makes sense to me. Ie. we can pick a subcover to cover the interval from 0 to 1/N, which contains infinitely many points, and then we only have finitely many " ie. 1, 1/2, 1/3, 1/4, 1/5,....1/(N-1)" left over to cover.
So to keep this subcover arbitrary in my argument I should never mention any specific subcover, I mean I should never say choose the collection as a subcover for etc. Correct? I should just say "a subcover can be chosen to cover the points in this interval: " etc.