# Thread: Real Analysis - Sequences 3

1. ## Real Analysis - Sequences 3

Need counter-example I believe

2. Originally Posted by Phyxius117

Need counter-example I believe
a) What about $\displaystyle x_n=(-1)^n,y_n=(-1)^{n+1}+\frac{1}{n}$

3. Originally Posted by Phyxius117

Need counter-example I believe
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.

4. Originally Posted by Drexel28
a) What about $\displaystyle x_n=(-1)^n,y_n=(-1)^{n+1}+\frac{1}{n}$
Originally Posted by redsoxfan325
d.) Assuming this sequence is in the metric space $\displaystyle (M,d)$, 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.
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 $\displaystyle L-L'=\lim\text{ }\{x_n+y_n\}-\lim\text{ }x_n=\lim\text{ }\{x_n+y_n-x_n\}=\lim\text{ }y_n$