Completeness

[@blackpenredpen]

No wonder I never feel my life is completed...

[@andrewgrebenisan6141]

Clickbait perfection

[@Jaojao_puzzlesolver]

My life is complete.

[@Bran4901]

Ah, but does your life have limits?

[@NotoriousSRG]

I love that you never pronounce “Cauchy” the same way twice lol

[@punditgi]

Such an eye opening explanation!

[@vadimpavlov6037]

I think the final step relies on the fact that R contains all its lim sups and lim infs, for otherwise the same argument would apply to Q

[@hauntedmasc]

This video on completeness is a (Dedekind) cut above the rest!

[@rajivdixitbhaiifollower5055]

Oooo yes! Now my life is completed

[@arijeetsingh9444]

This channel probably has the lowest dislike/like ratio! Way to go Dr. Peyam.

[@toaj868]

Can you please explain where this proof breaks down if our range is restricted to Q? Is it that the Limsup and Liminf do not exist?

[@drpeyam]

Pretty much, since a closed subset of a complete space is complete

[@rajivdixitbhaiifollower5055]

Great sir ! Well done

[@MuPrimeMath]

lol that title

[@dmytro_shum]

Sequence: sum 1/k (harmonic series) for k from 1 to n is not bounded. Is it a Cauchy sequence?

[@dmytro_shum]

@@chemhwa I'm asking the author of the video

[@dstigant]

No because it’s not bounded. Consecutive terms (Sn and Sn-1) will be arbitrarily close, but arbitrarily distant terms will not be arbitrarily close. That is , for any N and ep, there exists another integer larger than N, say M, Such that S(M)-S(N) > ep.

[@dmytro_shum]

@@dstigant Another guy, which cant read ))) I am write that to interest him to make a video about that, because it is much more interesting sequence to use as example

[@plaustrarius]

Yess! It is completed!!

[@toaj868]

A space can be incomplete if we can get an element outside that space in the limit. Is there any other way in which spaces can be incomplete?

[@drpeyam]

Any metric space can be completed, so no

[@pocojoyo]

Dr Peyam, is this part of your Elementary Real Analysis course ? 1 or 2 ?

[@drpeyam]

Yes, and 1

[@pocojoyo]

@@drpeyam Thanks !

[@B_A-tr]

So large amounts of people are cauchy

[@dgrandlapinblanc]

Ok. Cool way. Thank you very much.

[@f3ynman44]

Where was this video when I took my Analysis Exam one month ago!? Still a good video though! 👍

[@henrikfischbeck7198]

ure saying supremum and max as if they where the same. Also u say Q has holes, but then u say Z is complete... if it was lim min = lim max i would undetrstand but.. Uhm?... liminf(q)=sqrt(2)=limsup(q)...? Imagine a sequence of rationals getting closer and closer to sqrt2... the two numbers next to sqrt2 are rational ( because between two rationals is an irrational. and vice versa) so a sequence of rationals alternating around sqrt2 getting closer and closer have liminf(q)=sqrt(2)=limsup(q) ? like Q is dense in R so the smallest distance d(q,sqr2). but not if we change supp and inf for min and max..?

[@shiina_mahiru_9067]

I have already forgotten the entire proof from my analysis class 😂

[@Galileo2pi]

Brilliant as always, mate, just on my birthday, thanks

[@late7245]

■ = the end

[@coreymonsta7505]

it's not surprising after you see it's just about technicalities regarding where elements lie

[@vladimirbenitocardenas6012]

This is an hard idea

[@gabrielmendozallanes7323]

UC Irvine is not complete without Dr. Peyam

[@drpeyam]

Awww, I miss you!!!

[@jkid1134]

You're not building up to p-adics or something, are you? :)

[@carloshurtado8723]

Completitud

[@sdonic]

limb soup anyone?