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.
@@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
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..?
No wonder I never feel my life is completed...
Clickbait perfection
My life is complete.
Ah, but does your life have limits?
I love that you never pronounce “Cauchy” the same way twice lol
Such an eye opening explanation!
Oooo yes! Now my life is completed
This channel probably has the lowest dislike/like ratio! Way to go Dr. Peyam.
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
This video on completeness is a (Dedekind) cut above the rest!
Great sir ! Well done
lol that title
Yess! It is completed!!
Brilliant as always, mate, just on my birthday, thanks
So large amounts of people are cauchy
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?
Pretty much, since a closed subset of a complete space is complete
Sequence: sum 1/k (harmonic series) for k from 1 to n is not bounded. Is it a Cauchy sequence?
@@chemhwa I'm asking the author of the video
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.
@@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
Ok. Cool way. Thank you very much.
Dr Peyam, is this part of your Elementary Real Analysis course ? 1 or 2 ?
Yes, and 1
@@drpeyam Thanks !
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?
Any metric space can be completed, so no
I have already forgotten the entire proof from my analysis class 😂
Where was this video when I took my Analysis Exam one month ago!?
Still a good video though! 👍
■ = the end
it's not surprising after you see it's just about technicalities regarding where elements lie
UC Irvine is not complete without Dr. Peyam
Awww, I miss you!!!
This is an hard idea
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..?
You're not building up to p-adics or something, are you? :)
Completitud
limb soup anyone?