Proof: Cauchy Sequences are Bounded | Real Analysis

We prove that every Cauchy sequence is bounded. We previously discussed what the Cauchy criterion and Cauchy sequences are, and proved that a sequence is Cauchy. We are on our way to proving a sequence converges if and only if it is Cauchy, but to help us in doing that, we will first prove that Cauchy sequences are bounded in today's real analysis video lesson.
To do this, we use the definition of Cauchy sequences to then use our useful absolute value inequality equivalence to establish a bound on terms after a certain point. Then, a max and min will clean up the rest.
What are Bounded Sequences: • What are Bounded Seque...
Proof Convergent Sequence is Bounded: • Proof: Convergent Sequ...
Proof of the Absolute Value Inequality Equivalence: • Proof: A Useful Absolu...
Intro to Cauchy Sequences: • Intro to Cauchy Sequen...
Proof Convergent Sequences are Cauchy: • Proof: Convergent Sequ...
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[@avtaras]

Crystal clear explanation. This is my go-to channel for Analysis.

[@PraveenYadav-wt3hd]

Watched all your Real Analysis videos.Best on TH-cam

[@WrathofMath]

Thanks a lot, Pooja! There will be more coming! If you're looking for some less intense math in the meantime, check out my newest video on an infamous shape! th-cam.com/video/uEJQubc0PPU/w-d-xo.html

[@Kevbotxd]

Wrath you came in clutch for me again today brother. Cauchy Definition and bounded proof was the first question on my 2nd midterm today. I'm passing this class because of you! 🎉😂

[@WrathofMath]

Because of me and because of all your hard work! And hopefully some help from your professor haha. Thanks for watching and I hope you'll continue to find the lessons helpful!

[@kabirbhattacharyya9014]

Thank you!!!! You're an absolute lifesaver

[@WrathofMath]

Glad to help! Thanks for watching!

[@yihanghu181]

Hi. Why L

[@physicsfaith]

We know for n>N that the terms get really close together so that for N>=1+ a_(N+1), the terms are smaller than that number. What about the a_n terms smaller than N? We have to account for them individually because some of them may be bigger than 1+a_(N+1). So we make a set of all the terms up to a_N and the last term as well. We know that the max of this set U is the upper bound of all the terms. The same goes for L but flipped for the lower bound.

[@physicsmee1]

Love you bro I am from india .......keep it doing.
You solved my confusion

[@WrathofMath]

Thanks for watching!

[@martin_skachkov6016]

Hey, thank you so much for this helpful video :) ! I don't know if here is the right place to ask you but do you have a video proving if a function is differentiable then its derivative is continuous?

[@WrathofMath]

Thanks for watching and for the question! Check out this link showing an example that implication does not hold: calculus.subwiki.org/wiki/Derivative_of_differentiable_function_need_not_be_continuous

[@martin_skachkov6016]

@@WrathofMath Thank you 🙂

[@siriuss_]

Thanks a lot, ah i spent so much time staring at my textbook's skipped lines discussion of this bounded sequence 🤦

[@Leyla-et9nh]

U R MY HERO

[@WrathofMath]

Glad to help! Thanks for watching and let me know if you ever have any video requests! If you're looking for more analysis, check out my playlist:
th-cam.com/play/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli.html

[@gaaraofddarkness]

isnt this L and U dependent on 𝜀=1, so how can we be sure for all 𝜀??

[@physicsfaith]

You could replace 1 with epsilon. Same thing.

[@nicksullivan4203]

Why is this true in all metric spaces. How can the norm of -1< am - an+1 < 1. for example if you use the discrete metric (1 if they are the same and 0 if the two values are not the same) then this just doesn't work right or am I missing something?

[@WrathofMath]

Thanks for watching and good question, in that example I don't think there could be a Cauchy sequence. If the terms of a sequence are eventually constant, then all the distances will be 1, and thus it can't be Cauchy, and if the terms are not eventually constant, the distance between any number and itself will always be 1, thus still failing to be Cauchy. Thus, it has no bearing on the result we discuss in the video.

[@nicksullivan4203]

@@WrathofMath ahh I see. Does this still work in metrics where the absolute value is not the norm. For example, L inf or ||•||2?