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$\begingroup$ The keyword here is "Bolzano-Weierstrass property." An easy example of a space which lacks this property is $(0, 1)$. An easy example of a space which lacks this property is $(0, 1)$. (Note that $(0, 1)$ is homeomorphic but not bi-Lipschitz to $\mathbb{R}$ and that convergence is a topological notion but boundedness is a metric one.) $\endgroup$
Bolzano-Weierstrass theorem (complex case) Ask Question Asked 9 years, 6 months ago. Modified 9 years, 6 ...
Dec 26, 2022 · If you know about this sort of thing, you might like to learn that Bolzano-Weierstrass is true in any complete metric space where bounded subsets are always totally bounded. Sequential compactness (essentially this is Bolzano-Weierstrass) is equivalent to compactness which is further (generalised Heine-Borel) equivalent to completeness and total boundedness (in Euclidean space, that is just closed and bounded).
$\begingroup$ Thanks for the green tick! I was just writing the following: "think of the sequence as a train line with infinitely many stops: Bolzano-Weierstrass tells you that there is an express train that may miss some of the stops but has a limit point at a final destination.
Oct 6, 2018 · 9. Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true. That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.
Apparently, it is not within the scope of elementary textbooks on real analysis to include proofs of all these equivalences. Among these results, the Heine-Borel theorem and the Bolzano-Weierstrass theorem are of fundamental importance in applications and generalization to a wider framework of topological spaces.
Jan 30, 2018 · The wikipedia proof of Bolzano Weierstrass theorem. 1. Question on proof of the Bolzano-Weierstrass ...
Feb 25, 2012 at 17:51. The Bolzano-Weierstrass theorem states that each bounded sequence has a convergent subsequence. It follows that an unbounded sequence has at least one diverging subsequence. – wjmolina. Feb 25, 2012 at 17:53. The cosines stay between −1 − 1 and 1 1, so you could not have found one of the subsequences you describe.
Proof : Bolzano Weierstrass theorem. As part of the complete proof the professor gave he proved this implication: Let A ⊂ R and every sequence (an)n∈N in A has at least one accumulation point in A. That implies, A is bounded and closed. The proof he gave ( only the part where he proves, that A is bounded) : A is bounded: Let's assume ( for ...
Nov 7, 2017 · A normed vector space satisfies the Bolzano-Weierstrass property (i.e. any bounded sequence has a convergent subsequence) if and only if it is of finite dimension. This means there is a counterexample in any infinite dimensional normed vector space.
