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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Then S S is a compact topological space with the induced topology precisely if it is complete and totally bounded with the induced metric. Cantor spaceMandelbrot space. This gives the next version:.
A generalisation applies to all metric spaces and even to uniform spaces. According to Wikipediathe theorem was first proved by Pierre Cousin in I would also like to point out that his approach is odd, and you would be better reading something like Rudin’s principles of Mathematical Analysis if you’re learning this stuff for the first time though sometimes reading Hardy’s book is nice for a change.
Warsaw circleHawaiian earring space. Let S S be a uniform space. Continuing in like manner yields a decreasing sequence of nested n -boxes:.
real analysis – What does the Heine-Borel Theorem mean? – Mathematics Stack Exchange
We need to show that thheorem is closed and bounded. Kris 1, 8 This sequence is Cauchy, so it must converge to some limit L. It turns out that this is all we need: S S is compact. Let us define a sequence x k such that each x k is in T k.
Assume that the set of admissible values were an open interval [ 0g [0,g.
The restrictions of these to S S hence form an open cover of the subspace S S. It is thus possible to extract from theoerm open cover C K of K a finite subcover.
Indeed, the intersection of the finite family of sets V U is a neighborhood W of a in R n.
By the assumption that S S is compact, there is then a finite subset of n n -tuples of integers such that the corresponding n n -cubes still cover S S. We could also try to generalise Theorem to subspaces throrem other metric spaces, but this fails: 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.
All Montel spaces have the Heine-Borel property as well. This is by no means a modern statment of the Heine-Borel Theorem. Sign up using Facebook. Hence, theorej accumulation point of S is in Sso S is closed.
Brouwer’s fixed point theorem. S S is closed and bounded. S is closed and bounded S is compactthat is, every open cover of S has a finite subcover.
Since closed subspaces of compact spaces are compact this implies that S S is compact. If a set S in R n is bounded, then it can be enclosed within an n -box.
I am working through Hardy’s Course of Pure Mathematics ed. Sign up or log in Sign up using Google.