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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Sign up using Email and Password. Thanks it makes sense now! By using this site, you agree to the Terms of Use and Privacy Policy. One example is the existence of a Pareto efficient allocation. Mathematics Stack Exchange works best with JavaScript enabled. There are different important equilibrium concepts in economics, the proofs of the existence of which often require variations of the Bolzano—Weierstrass theorem.

Bolzsno fifty years later the result was identified as significant in its own right, and proved again by Weierstrass. It doesn’t matter, but it’s a neater proof to say weiersttrass the left hand one. If that’s the case, you can pick either one and move on to the next step.

Post Your Answer Discard 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. Does that mean this proof only proves that there is only one subsequence that is convergent?

We continue this process infinitely many times. Views Read Edit View history. Post as a guest Name. I just can’t convince myself to accept this part. The proof is from the book Advanced Calculus: This page was last edited on 20 Novemberat I know because otherwise you wouldn’t have thought to ask this question. Indeed, we have the following result.

It follows from the monotone convergence theorem that this subsequence must converge.

Theorems in real analysis Compactness theorems. It has since become an essential theorem of analysis. In mathematics, specifically in real analysisthe Bolzano—Weierstrass theoremnamed after Bernard Bolzano and Karl Weierstrassis a fundamental result about convergence in a finite-dimensional Euclidean space R n.

The theorem states that each bounded sequence in R n has a convergent subsequence. Thus we get a sequence of nested intervals. Now, to answer your question, as others have said and you have said yourselfit’s weierstras possible that both intervals have infinitely many elements from the sequence in them.

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. Sign up or log in Sign up using Google. Since you can choose either one in this case, why not always just choose the left hand weierstras Retrieved from ” https: From Wikipedia, the free encyclopedia.

Thank you for the comments! 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.

Home Questions Tags Users Unanswered. There is also an alternative proof of the Bolzano—Weierstrass theorem using nested intervals. Because we halve the length of an interval at each step the limit of the interval’s length is zero.

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. I am now satisfied and convinced, thank you so much for the explanation!

We will use peoof method of interval-halving introduced previously to prove the existence of least upper bounds. 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.

Michael M 2, 6 Email Required, but never shown. 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. It was actually first prkof by Bolzano in as a lemma in the proof of the intermediate value theorem.

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. Sign up using Facebook. Because each sequence has infinitely many members, there must be at least one subinterval which contains infinitely many members.

To show existence, you just have to show you can find one. Help me understand the proof for Bolzano-Weierstrass Theorem!

I boxed the part I didn’t understand. By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Your brain does a very good job of checking the details.