11:56 Consider a diverging sequence X_n, defined by Xn= n if n is odd and Xn= 1 if n is even. Here the sequence Xn is diverging but the subsequence X_(2n) is converging.
Not need of prove for option 1 is complete only one type of chochy sequence in which nth term become constant so it is complete. Option 2 sequence become converse on 0 not in natural no set so it is not complete. Option 3 every seuence should chauchy for totally bounded but when nth term is not constant it is divergent not chochy not totally not bounded Option 4 all types of this sequence always convergent always chauchy so totally bounded
Sir, in general it is not true dat subsequence of a divergent sequence is divergent.But you stated that since x_n is divergent all of its subsequence are divergent. why?
Thankue for asking a good question we always appreciate that Please think on this Sequenece is not convergent does not means sequence is divergent For example Xn = (-1)^n is not a divergent sequence
so complicated to prove totally boundedness but you explain nicely. Is there any equivalent definition of totally boundedness which is easy to understand?
11:56 Consider a diverging sequence X_n, defined by Xn= n if n is odd and Xn= 1 if n is even. Here the sequence Xn is diverging but the subsequence X_(2n) is converging.
If a sequence is divergent that does not mean always that all of it's subsequence also will be divergent.
= (-1)^n.
Totally bounded implies bounded. Now X1 is not bounded with the metric d1, so X1 is not totally bounded.
Not need of prove for option 1 is complete only one type of chochy sequence in which nth term become constant so it is complete.
Option 2 sequence become converse on 0 not in natural no set so it is not complete.
Option 3 every seuence should chauchy for totally bounded but when nth term is not constant it is divergent not chochy not totally not bounded
Option 4 all types of this sequence always convergent always chauchy so totally bounded
In option 4 : (1/n) is not convergent in d2.am I right.pls explain.pls sir
Because in (1/n) : d2( 1/6,1/7)=-1which is not less than epsilon .pls explain about my view.pls sir
Sir, in general it is not true dat subsequence of a divergent sequence is divergent.But you stated that since x_n is divergent all of its subsequence are divergent. why?
Thankue for asking a good question we always appreciate that
Please think on this
Sequenece is not convergent does not means sequence is divergent
For example Xn = (-1)^n is not a divergent sequence
Really nice sir
so complicated to prove totally boundedness but you explain nicely. Is there any equivalent definition of totally boundedness which is easy to understand?
See in my comment totally bounded definition
So confusing
What you have found confusing rimpa the question itself or the the way it is being taught . You can comment your querry rimpa if you have doubt
See in my comment explanation