Students sometimes struggle with the Heine-Borel Theorem; the authors certainly did the first time it was presented to them. This theorem can be hard to. Weierstrass Theorem and Heine-Borel Covering Theorem. Both proofs are two of the most elegant in mathematics. Accumulation Po. Accumulation Points. Heine-Borel Theorem. October 7, Theorem 1. K C Rn is compact if and only if every open covering 1Uαl of K has a finite subcovering. 1Uα1,Uα2,,Uαs l.

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Warsaw circleHawaiian earring space. Lemma closed interval is compact In classical mathematics: Central to the theory was the concept of uniform continuity and the theorem stating that norel continuous function on a closed interval is uniformly continuous.

The restrictions of these to S S hence form an open cover of the thekrem S S. A generalisation applies to all metric spaces and even to uniform spaces. Hence what remains is to show that S S is bounded.

### Heine–Borel theorem – Wikipedia

The proof above applies with almost no change to showing that any compact subset S of a Hausdorff topological space X is closed in X. If S is compact but not closed, then it has an accumulation point a not in S. Note that we say a set of real numbers is closed if every convergent sequence in that set has its limit in that set. Peter Gustav Lejeune Dirichlet was the first to prove this and implicitly he used the existence of a finite subcover of a given open cover of a closed interval in his proof.

S S is compactS S is closed and bounded. Continuing in like manner yields a decreasing sequence of nested n -boxes:. The history of what today is called the Heine—Borel theorem starts in the 19th century, with the search for solid foundations of real analysis. This sequence is Cauchy, so it must converge to some limit L.

By the property above, it is enough to show that T 0 is compact. From Wikipedia, the free encyclopedia. In fact it holds much more generally than for subspaces of a cartesian space:.

I am confused as to what this theorem is actually saying. Then there exists an infinite open cover C of T 0 that does not beine any finite subcover. Thus, T 0 is compact. By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that your continued use of the website is subject to these policies.

Since closed subspaces of compact spaces are compact this implies that S S is compact.

S S is compact. I’ll disect what Hardy discusses below.

Sign up using Facebook. Proof Since all the closed intervals are homeomorphic it is sufficient to show the statement for [ 01 ] [0,1]. Mathematics Stack Exchange works best with JavaScript enabled. Likewise, the sides of T 1 can be bisected, yielding 2 n sections of T 1at least one of which must require an infinite subcover of C.

It’s easy to prove that S S is closed precisely if it is a complete metric space as with the induced metricand similarly S S is bounded precisely if it is totally bounded. Then S S is a compact topological borep with the induced topology precisely if it is complete and totally bounded with the induced metric. By using this site, you agree to the Terms of Use and Privacy Policy.

I was trying to learn heime from Hardy a while ago, and Hsine found it quite difficult due to his old fashioned way of speaking.

All Montel spaces have the Heine-Borel property as well. This theorem refers only to uniform properties of Borep Sand in fact a further generalistion is true:. We do so by observing that the alternatives lead to contradictions: Email Required, but never shown.

## Heine–Borel theorem

Ua contradiction. Theorems in real analysis General topology Properties of topological spaces Compactness theorems. Since all the closed intervals are homeomorphic it is sufficient to show the statement for [ 01 ] [0,1].