Complete Metric Space | Lecture 3 | Every convergent sequence is bounded

[@educationfreely6122]

Nice teaching sirji. God bless you

[@shreyaghosh729]

Thank u sir 🙏 ye videos se hard se hard topic ko v jaldi se samjh aa jati hai ...

[@ranjankhatu]

Thank you

[@pradeepbaghel838]

Nice teaching method

[@ranjankhatu]

Thank you

[@souravpaul_per]

Thanks sir

[@canearth5927]

My favorite Pq

[@ranjankhatu]

Thank you

[@ranjankhatu]

Thank you

[@sakshipawar4920]

thank youu sir

[@prernakumari1431]

Thank you sir

[@ranjankhatu]

Most welcome

[@keshavlalit1555]

Sir we can take any arbitrary epsilon na ,not necessarily 1 ?

[@ranjankhatu]

For any epsilon, definition works. So it works for 1 also. And we did the same.

[@keshavlalit1555]

@@ranjankhatu if we don't take 1 , can we do by directly taking epsilon

[@ranjankhatu]

@@keshavlalit1555 to prove the sequence bounded, we need to prove modulus of all elements are less than on fixed number. That's why we took epsilon.

[@322-pinkidhobi9]

let (an, bn, cn) be a sequence in( R^3, d2) and (p, q, r) €R^3 then( an, bn, cn) convergent to (p, q, r) in( R^3, d2) iff (an )convergent to p and (bn) cong. q and( cn) convergent to r in (R, | |) how to prove this question

[@ranjankhatu]

I hope my following video will help you to write its proof
ua-cam.com/video/wjRj8Oifjqg/v-deo.html