# BOLZANO WEIERSTRASS THEOREM PROOF PDF

Two other proofs of the Bolzano-Weierstrass Theorem. We prove the result: If \$ \ mathbb{X} = \{x_n: n \in \mathbb is a sequence of real numbers. Theorem. (Bolzano-Weierstrass). Every bounded sequence has a convergent subsequence. proof: Let be a bounded sequence. Then, there exists an interval. The proof doesn’t assume that one of the half-intervals has infinitely many terms while the other has finitely many terms; it only says that at least one of the halves . Author: Tur Akinokus Country: Mexico Language: English (Spanish) Genre: History Published (Last): 10 October 2008 Pages: 367 PDF File Size: 15.82 Mb ePub File Size: 17.80 Mb ISBN: 783-4-54901-167-5 Downloads: 99732 Price: Free* [*Free Regsitration Required] Uploader: Baramar It has since become an essential theorem of analysis.

### Bolzano–Weierstrass theorem – Wikipedia

The Bolzano—Weierstrass theorem allows one to prove that if the set of allocations is compact weirstrass non-empty, then the system has a Pareto-efficient allocation. The theorem states that each bounded sequence in R n has a convergent subsequence.

By using this site, you agree to the Terms of Use and Privacy Policy. Home Questions Tags Users Unanswered. Theorems in weiestrass analysis Compactness theorems. Bolzno it makes sense now! Since you can choose either one in this case, why not always just choose the left hand one? The proof doesn’t assume that one of the half-intervals has infinitely many terms while the other has finitely many terms; it only says that at least one of the halves has infinitely many terms, and one can be chosen arbitrarily.

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I just can’t convince myself to accept this part.

Retrieved from ” https: I am now satisfied and convinced, thank you so much for the explanation! Indeed, we have the following result. Views Read Edit View history.

Now, to answer your question, as others have said and you have said yourselfit’s entirely possible weirrstrass both intervals have infinitely many elements from the sequence in them.   I know because otherwise you wouldn’t have thought to ask this question. Does that mean this proof only proves that there is only one subsequence that is convergent? It doesn’t matter, but it’s a neater proof to say “choose the left hand thforem. To show existence, you just have to show you can find one.

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Some fifty years later weierstrass result was identified as significant in its own right, and proved again by Weierstrass. In fact, general topology tells us that a metrizable space is compact if and only if it is sequentially compact, so that the Bolzano—Weierstrass and Heine—Borel theorems are essentially the same.