Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

I am wondering why the "creeping along" proof of Heine-Borel theorem doesn't work for open subsets. I have adapted Spivak's proof in "Calculus on Manifolds" here, but I can't seem to find where I am wrong:

The theorem for open sets would be:

The open interval (a, b) is compact.

"Proof":

Let O be an open cover of (a,b).

Let A = {x : a <= x <=b and (a,x) is covered by some finite number of open sets in O}.

Then A is clearly bounded above by b.

We want to show that b \in A, which is done by proving that \alpha, the supremum of A, is in A and that \alpha = b.

Since O is a cover, \alpha \in U for some U in O. Then all points in some interval to the left of \alpha are also in U. Since \alpha is the least upper bound of A, there are x_i and x_j in this interval such that x_i, x_j \in A and x_i < x_j. Thus (a, x_j) is covered by some finite number of open sets of O, while (x_i, \alpha) is covered by the single set U. Hence (a,\alpha) is covered by a finite number of open sets of O, and \alpha \in A.

Now, suppose instead that \alpha < b. Then there is a point y between \alpha and b such that (\alpha, y) is a subset of U. Since \alpha \in A, the interval (a, \alpha) is covered by finitely many open sets of O, while (\alpha, y) is covered by U. Hence y \in A, contradicting the fact that \alpha is an upper bound of A.

I would be grateful to anyone who finds mistakes in the above paragraphs.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Why the proof of Heine-Borel theorem doesn't work for open sets.

Loading...

Similar Threads for proof Heine Borel |
---|

B Proof of a limit rule |

B Proof of quotient rule using Leibniz differentials |

B Don't follow one small step in proof |

**Physics Forums | Science Articles, Homework Help, Discussion**