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 .

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There are different important equilibrium concepts in economics, the proofs of the existence of which often require variations of the Bolzano—Weierstrass theorem.

An allocation is a matrix of consumption bundles for agents in an economy, and an allocation is Pareto efficient if no change can be made to it which makes no agent worse off and at least one agent better off here rows of the allocation matrix must be rankable by a preference relation.

Moreover, A must be closed, since from a noninterior point x in the complement of Aone can build an A -valued sequence converging to x. This form of the theorem makes especially clear the analogy to the Heine—Borel theoremwhich asserts that a subset of R n is compact if and only if it is closed and bounded.

Home Questions Tags Users Unanswered. 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 boozano.

In mathematics, specifically in real analysisthe Bolzano—Weierstrass theoremnamed after Bernard Bolzano and Karl Weierstrassis a fundamental result about weiefstrass in a finite-dimensional Euclidean space R n. Since you can choose either one in this case, why not always just choose the left hand one?

Michael M 2, 6 There is also an alternative proof of the Bolzano—Weierstrass theorem using nested intervals. By using this site, you agree to the Terms of Use and Privacy Policy.

From Wikipedia, the free encyclopedia. This page was last edited on 20 Novemberat It has since become an essential theorem of analysis. By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Post Your Answer Discard By clicking “Post Your Answer”, you acknowledge that you have read weoerstrass updated terms of serviceprivacy policy and cookie policyand that your continued use of the website bbolzano subject to these policies.

Help me understand the proof for Bolzano-Weierstrass Theorem! I just can’t convince myself to accept this part. We continue this process infinitely many times.

Now, to answer your question, as seierstrass have said and you have said yourselfit’s entirely possible that both intervals have infinitely many elements from the sequence in them. Post as a guest Name. I am now satisfied and convinced, thank you so much for the explanation! Some fifty years later the result was identified as significant in its own wsierstrass, and proved again by Weierstrass. The theorem states that each bounded sequence in R n has a convergent subsequence.

To show existence, you just have to show you can find one. Indeed, we have the following result.

By clicking “Post Your Answer”, you acknowledge that you have read our updated terms of serviceprivacy policy and cookie policyand that weirrstrass continued use of the website is subject to these policies. It follows from the monotone convergence theorem that this subsequence must converge. It was actually first proved by Bolzano in as a lemma in the proof of the intermediate value theorem. Mathematics Stack Exchange works wieerstrass with JavaScript enabled.

If that’s the case, you can pick either one and move on to the next step.

Thanks it makes sense now! Retrieved from ” https: Views Read Edit View history.

Thank you for the comments! The Bolzano—Weierstrass theorem allows one to prove that if the set of allocations is compact and non-empty, then the system has a Pareto-efficient allocation. Your brain does a very good job of checking the details. We will use the method of interval-halving introduced previously to prove the existence of least upper bounds. Thus we get a sequence of nested intervals. 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.

I know because otherwise you wouldn’t have thought to ask this question. Because each sequence has infinitely many members, there must be at least one subinterval which contains infinitely many members. Sign up or log in Sign up using Google. Sign up using Email and Password. Email Required, but never shown. Sign up using Facebook.

It doesn’t matter, but it’s a neater proof to say “choose the left hand one. I boxed the part I didn’t understand. Does that mean this proof only proves that there is only one subsequence that is convergent?

Suppose A is a subset of R n with the property that every sequence in A has a subsequence converging to an element of A. One example is the existence of a Pareto weierstrzss allocation.

Theorems in real analysis Compactness theorems.