http://i10.photobucket.com/albums/a1...7/problem8.jpg

Please help!!

Need counter-example I believe

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- Feb 16th 2010, 12:53 PMPhyxius117Real Analysis - Sequences 3
http://i10.photobucket.com/albums/a1...7/problem8.jpg

Please help!!

Need counter-example I believe - Feb 16th 2010, 12:57 PMDrexel28
- Feb 16th 2010, 02:14 PMredsoxfan325
d.) Assuming this sequence is in the metric space $\displaystyle (M,d)$, if $\displaystyle \{b_n\}$ converges, then it is bounded. ($\displaystyle \exists x_0\in M, R\in\mathbb{R}$ such that $\displaystyle d(x_0,b_n)<R~\forall n$)

So $\displaystyle d(a_n,x_0)-R<d(a_n,b_n)~\forall n$. (Triangle Inequality)

But since $\displaystyle \{a_n\}$ is unbounded, so is $\displaystyle d(a_n,x_0)$, because $\displaystyle x_0$ is a fixed point. Therefore $\displaystyle d(a_n,b_n)$ is unbounded as well.

@ Drexel: $\displaystyle x_n$ needs to converge. - Feb 16th 2010, 02:18 PMDrexel28
a) is redonkulous then. Let $\displaystyle \lim\text{ }\{x_n+y_n\}=L$ and $\displaystyle \lim\text{ }x_n=L'$. Since they both converge we see that $\displaystyle L-L'=\lim\text{ }\{x_n+y_n\}-\lim\text{ }x_n=\lim\text{ }\{x_n+y_n-x_n\}=\lim\text{ }y_n$