Talk:Heine–Borel theorem
From Wikipedia, the free encyclopedia
I removed the following proof from the main page:
[edit] Proof
Here, we consider the version for the real numbers, as stated above.
[edit] ->
If a subset of the real numbers is compact, then it is bounded.
Let X be a compact subset of the real numbers.
Define a collection of open sets O_n to be the open interval (-n, +n). These sets cover X. Some finite sub-collection must also cover X. This finite sub-collection is bounded, and so X is bounded.
If a subset of the real numbers is compact, then it is closed.
Suppose (for a contradiction) that it were not closed. Then, there would be a sequence of number x_n in X such that x_n converges to a point x not in the set X.
Define the sequence of open sets O_n where O_n = R - [x - (1/n), x+ (1/n)], i.e. the whole line minus a small closed interval around x. O_n is open and the collection covers X. Since X is compact, a finite subcover exists. So, every point in X is at least 1/n away from x (n fixed). So, a sequence in X cannot converge to a point outside of X. Thus, X is closed.
[edit] <-
We are given a closed and bounded subset X of R. We are given an open cover {O} of X. We need to find a finite subcover.
Step 1: Reduce the open cover to a countable subcover. For each rational number r, pick one open set in O_r that contains r. We use the axiom of choice here. Since X is closed, this subcollection also covers X.
Step 2: We now have a countable subcover {O_n}. Suppose that there is no finite subcover. Consider P_n = the union of the first n members of O_n. Each P_n is open, and does not cover X. So, there is some point x_n in X - P_n. Since X is closed and bounded, every sequence in X has some convergent subsequence (A proof of this should also be given; in fact, some books list this as the definition of compactness). This convergent subsequence, call it y_n, converges to a point y. However, since X is closed, y is also in X. So, y is covered by some O_n, and y is in some P_n. This is a contradiction, so there must be some finite subcover.
We still need to prove that, any sequence in a closed and bounded set has a convergent subsequence. Since the set X is bounded, it is bounded by a closed interval [-n, +n]. Now, one of the two sub-intervales [-n, 0] and [0, +n] must contain an infinite number of points in the sequence. Repeat this process, dividing the interval in half, finding a subsequence contained in that half. Eventually, this yields a convergent subsequence. The axiom of choice is again required.
Important point: the Axiom of Choice was used.
Since this theorem is important in analysis, the Axiom of Choice is also important in analysis.
The first half of the proof is correct, but the second half is not. Step 1, the reduction to a countable subcover, is not correct: the constructed countable subcover need not cover all of X. The Axiom of Choice is in fact not needed for the proof. AxelBoldt 18:52 Oct 4, 2002 (UTC)
What's a correct proof? Overall, I was expecting to see more proofs and examples on the Mathematics pages.
