Sequence Converges iff Every Subsequences Converge to the Same Limit | Real Analysis

[@WrathofMath]

Check out the full Real Analysis playlist for more! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli Lesson on subsequences: kzbin.info/www/bejne/gaK7YaCBmLuSaaM

[@ranimahassen6296]

You are literally the best analysis teacher around. So simple, direct and well-explained! Thank you!!!

[@chisae_]

Literally saved my life. I've been watching your Real Analysis playlist in preparation for my exam tomorrow. Thank you so much for the videos!! Very helpful.

[@WrathofMath]

So glad to help - thanks for watching! Good luck on the exam!

[@michaelyu2015]

Imagine the difficulty of learning this first handed with English being your second language, it's like listening to a puzzle.

[@WrathofMath]

I'm sure that's tremendously difficult, much respect for your effort! Let me know if you have any questions I can help clarify.

[@farhanniazi94]

You have some of the best videos on analysis. Every theorem follows a motivating example; every result is presented in the right order with enough justification. If you write a book i’ll pre-order instantly. Btw, at 2x speed you sound like Jesse Eisenberg

[@nonentity168]

Almost got lost in those notations. Thanks a lot for clarification.

[@WrathofMath]

Glad it helped!

[@sambitgarai3195]

I can’t visualise the part where k> N1 then k> N is assumed. Isnt N1 < N. theoretically, k< N

[@sethsullivan7031]

This was amazing!! Thank you!

[@manuelkarner8746]

7:13 loool that´s a great little proof

[@WrathofMath]

Haha, thanks Manuel! They don't come much easier than that! Basically "A is A, prove A is A".

[@ThePhysicsMathsWizard]

Very nice :) ! Is this the Bolzano - wierstrass theorem?

[@WrathofMath]

Thank you! This is not the Bolzano-Weierstrass theorem, but we'll be getting to that soon. In fact, I don't think we even use this result for the BW theorem. The BW theorem for sequences states that every bounded sequence has a convergent subsequence. Looking forward to making the proof video!

[@vaibhavoutat]

What stops us from taking finite subsequences, can't subsequences be finite? If not can you please point me to some resource mentioning the same, else if they can be finite how does this proof follow for finite subsequences.

[@abrilarellano2713]

does anybody know what's the name of the song at the end?, i really liked the video by the way :)

[@WrathofMath]

Thank you! The song at the end is 'Excited' by Crayon Angel/Vallow/Ben Watts. He is an elusive musician and unfortunately, with most of his music, has vanished from the internet. You can find a small amount of his music on this bandcamp page mountseldomrecords.bandcamp.com/ under the name "Vallow", but the song 'Excited' is no longer available. I have it on my computer though and would be happy to email it to you if you provide an email. I wish more people knew of him - he is an incredible songwriter!

[@abrilarellano2713]

​@@WrathofMath I just sent you an email, thanks! :)

[@WrathofMath]

I never received your email, try again and confirm you used the right address [e-mail]

[@kennethnavarro661]

Im so confused. If the sequence (a_n) converges to a then there exists some N in the naturals s.t. when n > N |a_n - a| < epsilon. Ok I get that, thats simple. But how does this imply every subsequence converges to a? If I take the first two values in the original sequence as my new subsequence for example (or any subsequence thats indices are less then N)?

[@kennethnavarro661]

Never mind I had a flawed view of what a subsequence was. I assumed that subsequences could be stopped after a limited amount of iterations but they actually go on forever. Forgive me!

[@WrathofMath]

Exactly, good job figuring it out!

[@vaibhavoutat]

@@kennethnavarro661 Where is it written that the subsequence can't be stopped can you please point me to the source. @WrathofMath

[@kushagrasinghal8209]

not to nitpick but ank need not be an infinite sequence or even have terms with indices bigger than K, so what you showed doesnt prove what you set out to prove. I guess I am nitpicking but its important to note.

[@NorceCodine]

This is the "freshman's proof" - can you come up with a proof that involves the Choice Function?