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Let X, | . |l be a normed vector space_ Give careful, precise definitions for each of the fol-lowing:The sequence Tn converges to 1 in the space Xb. Tn is a Cauchy ...

Question

Let X, | . |l be a normed vector space_ Give careful, precise definitions for each of the fol-lowing:The sequence Tn converges to 1 in the space Xb. Tn is a Cauchy sequence in the space XThe series Tn is convergent in the space X n=0The seriesTn is absolutely convergent in the space Xe: The space X is complete

Let X, | . |l be a normed vector space_ Give careful, precise definitions for each of the fol- lowing: The sequence Tn converges to 1 in the space X b. Tn is a Cauchy sequence in the space X The series Tn is convergent in the space X n=0 The series Tn is absolutely convergent in the space X e: The space X is complete



Answers

Cauchy Sequence A sequence $\left\{s_{n}\right\}$ is said to be a Cauchy sequence if and only if for each $\varepsilon>0,$ there exists a positive integer $N$ for which $$ \left|s_{n}-s_{m}\right|<\varepsilon \quad \text { for all } n, m>N $$ Show that every convergent sequence is a Cauchy sequence.

Hello. So he want to show that every convergent sequence is a koshi sequence. So here let's let a sub n B. A convergent sequence. And that the limit as N approaches infinity of a sub N B equal to L. Then, since a seven approaches L. So um for any greater than zero there exists a positive integer M. Such that we have the absolute value of a sub n minus L. Is then going to be less than such as absalon over to whenever we have that end is greater than M. So for an equal to M we then have the absolute value of a sub m minus one is less than epsilon over two. So then we have that the absolute value of a sub n minus a sub M is going to be equal to the absolute value of a sub m minus l minus a sub m Um -1. Which is going to be less than or equal to the absolute value of a sub. Um and -1 plus the absolute value of a sub m minus one which is going to be less than or equal to. Well epsilon over two plus epsilon over two, which is less than equal to epsilon. So and this is for all N for all and greater than or equal to M. So therefore by definition there that then shows that a sub end is going to be so, a sub N is a call. See they koshi sequence. Soco she was just about koshi pushy. Mhm. Yeah.


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