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The Bolzano-Weierstrass theorem Part 1

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We state and prove the Bolzano-Weierstrass theorem.
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Text Comments (23)
ajsdoa (1 month ago)
6:01 when you're talking, it sounds like you can just pick any x_k from the sequence in your subsequence with no particular pattern. As I've understood it, you couldn't just pick for instance {x_1,x_2,x_5,x_8,x_33,x_40,...}, just randomized numbers, I believe there has to be some kind of pattern like {x_2,x_4,x_6} for instance.
Alex Ware (1 month ago)
Very useful video!
Ankit Gothwal (2 months ago)
Thanks man. This video helped me a lot...keep making more of such videos...keep it up
Dimple Garg (3 months ago)
Waste of time
Shatakshi Sharma (3 months ago)
Sir if we take set of rationals between 0 and 1 this is also infinite and bounded subset of R but has no limit points....so does bolzano weirstras theorem fails in this case??
qwer1234 (7 months ago)
not entirely accurate, the set must be closed and bounded (ie. compact) for the weierstrass theorem to be applied. for example, 1<x<2 is a bounded set, with the sequence xk = {1.9,1.99,1.999,1.9999,1.99999,1.99999,1.9999999,....) being contained entirely within it, however no convergent subsequence can be obtained because it isn't closed.
mathIsART (7 months ago)
The way you explain math. It's mesmerizing. It's beautiful. I have no words.
peashooter (11 months ago)
This is totally irrelevant but you look quite like Freddie Highmore..
Robert Wilson III (1 year ago)
So basically, since the original sequence is bounded and the original sequence isn't convergent, you could just remove all the problem causing elements of the original sequence to therefore guarantee a convergent subsequence. You'd think textbook authors could just write what I've written above in their textbook plainly (since this is actually a simple concept), but they don't and consequently leave students confused. This really helps me out with that (1, 0) example. I was having trouble trying to understand how a bounded sequence could be divergent but this video cleared that up for me. Thanks.
Yutong chen (1 year ago)
Love you! Really clear! Thanks for sharing :)
Shiying He (2 years ago)
really helpful! thank you!!!
Noor Na (2 years ago)
Every convergent sequence is bounded, but not every bounded sequence is convergent
J Yu (5 months ago)
If a sequence is convergent, it is bounded. If bounded infinite set of real numbers contains an accumulation point A, then there exists a sequence in the set converging to A.
tolgahan21 (11 months ago)
Yes, I was just about to write that "Bolzano-Weierstrass Theorem: A bounded sequence of real numbers has a convergent subsequence" :)
Hugh Jones (2 years ago)
+Noor Na But every bounded sequence (in Rn) has a convergent subsequence. :-)
Nikola Tesla (2 years ago)
thanks, im from mexico. To my teacher i dont understand, he explain so fast and so abstrac some time XD
Parag Shah [USG] (2 years ago)
Ptolemy Pi (2 years ago)
Your videos are always very helpful.
Ahmad ElMoslimany (2 years ago)
Ruslan Jantayev (2 years ago)
Thanks !
Cool man, thanks!
Mohammed Alabboodi (3 years ago)
Thanks to explain that. Please can you put some examples!!!
Oscar Castillo (3 years ago)
Thank You! the idea of bounded and this theorem make sense to me now!

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