Real Analysis - Sequences 3

**Phyxius117** · February 16th 2010, 12:53 PM

Please help!!

Need counter-example I believe

**Drexel28** · February 16th 2010, 12:57 PM

Originally Posted by Phyxius117

Please help!!

Need counter-example I believe

a) What about $x_n=(-1)^n,y_n=(-1)^{n+1}+\frac{1}{n}$

**redsoxfan325** · February 16th 2010, 02:14 PM

Originally Posted by Phyxius117

Please help!!

Need counter-example I believe

d.) Assuming this sequence is in the metric space $(M,d)$ , if $\{b_n\}$ converges, then it is bounded. ( $\exists x_0\in M, R\in\mathbb{R}$ such that $d(x_0,b_n)<R~\forall n$ )

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

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

@ Drexel: $x_n$ needs to converge.

**Drexel28** · February 16th 2010, 02:18 PM

Originally Posted by Drexel28

a) What about $x_n=(-1)^n,y_n=(-1)^{n+1}+\frac{1}{n}$

Originally Posted by redsoxfan325

d.) Assuming this sequence is in the metric space $(M,d)$ , if $\{b_n\}$ converges, then it is bounded. ( $\exists x_0\in M, R\in\mathbb{R}$ such that $d(x_0,b_n)<R~\forall n$ )

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

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

@ Drexel: $x_n$ needs to converge.

a) is redonkulous then. Let $\lim\text{ }\{x_n+y_n\}=L$ and $\lim\text{ }x_n=L'$ . Since they both converge we see that $L-L'=\lim\text{ }\{x_n+y_n\}-\lim\text{ }x_n=\lim\text{ }\{x_n+y_n-x_n\}=\lim\text{ }y_n$

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