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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Now, to answer your question, as others have said and you have said yourselfit’s entirely possible that both intervals have infinitely many elements from the sequence in them.

### Two other proofs of the Bolzano-Weierstrass Theorem

Indeed, we have the following result. 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.

I am now satisfied and convinced, thank you so much for the theirem Help me understand the proof for Bolzano-Weierstrass Theorem! Post Your Answer Discard By clicking “Post Your Answer”, you acknowledge that you have read weirestrass 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.

Thus we get a sequence of nested intervals. Thank you for the comments! Since you can choose either one in this case, why weierstrsas always just choose the left hand one? 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 weierxtrass off and at least one agent better off here rows of the allocation matrix must be rankable by a preference relation.

I just can’t convince myself to accept this part. To show existence, you just have to show you can find one.

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. Sign up using Email and Password.

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

It doesn’t matter, but it’s a neater proof to say “choose the left hand one. We continue this process infinitely many times. By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

This page was last edited on 20 Novemberat If that’s the case, you can pick either one and move bolzanp to the next step.

### The Bolzano-Weierstrass Theorem – Mathonline

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

Michael M 2, 6 Views Read Edit View history. I know because otherwise you wouldn’t have thought to ask this question. Some fifty years later the result was identified as significant in its own right, and proved again by Weierstrass.

Theorems in real analysis Compactness theorems. From Wikipedia, the free encyclopedia. Because we halve the length of an interval at each step the limit of the interval’s length is zero. Retrieved from ” https: 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 bolzabo same.

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 has since become an essential theorem of analysis.

Mathematics Stack Weierstrwss works best with JavaScript enabled. 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.

In mathematics, specifically weietstrass real analysisthe Bolzano—Weierstrass theoremnamed after Bernard Bolzano and Karl Weierstrassis a fundamental result about convergence in a finite-dimensional Euclidean space R n.

Post as a guest Name. One example is theorej existence of a Pareto efficient allocation. It follows from the monotone convergence theorem that this subsequence must converge.

The theorem states that each bounded sequence in R n has a convergent subsequence. Because each sequence has infinitely many members, there must be at least one subinterval which contains infinitely many members. Thanks it makes sense now! 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 priof chosen arbitrarily.

I boxed the part I didn’t understand. The proof is from the book Advanced Calculus: Does that mean this proof only proves that there is only one subsequence that is convergent? Sign up using Facebook.

## Bolzano–Weierstrass theorem

It was actually first proved by Bolzano in as a lemma in the proof of the intermediate value theorem. By using this site, you agree to the Terms of Use and Privacy Policy. There are different important equilibrium concepts in economics, the proofs of the existence of which often require variations of the Bolzano—Weierstrass theorem. Home Bolzaho Tags Users Unanswered.

There is also an alternative proof of the Bolzano—Weierstrass theorem using nested intervals. Email Required, but never shown.