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Some help with the following would be great.

Let $(X,||\cdot||)$ be a normed space.

Let $(x_{n})_{n}$ and $(y_{n})_{n}$ be Cauchy sequences in $(X, D)$. Say also that $s_{n} = ||x_{n} + y_{n}||$.

Prove that $s_{n}$ is a Cauchy sequence in $(\mathbb{R}, |\cdot|)$.

Use the following:

$\large{|} \hspace{2pt} \normalsize||x-y||-||u-v|| \hspace{2pt}\large{|} \normalsize \le ||x-u|| + ||y-v||$

So I'm guessing we need to use the fact that $(x_{n})_{n}$ and $(y_{n})_{n}$ are Cauchy, then need to get it into the form of the hint.

But how?

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Come on, at least try to follow your nose and write down what you need to prove: the definition of $s_n$ being Cauchy. There is really only one thing to do. –  wildildildlife Feb 28 '12 at 23:55
    
To further expand on the previous comment: This kind of exercise is essentially just a matter of applying the definition. "Let $\varepsilon > 0$, then $|s_n-s_m| = \dotsb \leq \dotsb = \lVert x_n - x_m\rVert + \lVert y_n - y_m\rVert$. Since $(x_n)$ and $(y_n)$ are Cauchy there exists an $N \in \mathbb N$ s.t. $[\ldots]$, so for $n,m \geq N$ we have $|s_n - s_m| < \varepsilon$, i.e. $(s_n)$ is Cauchy $\quad\blacksquare$" –  kahen Feb 29 '12 at 0:17
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2 Answers 2

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Note that $$|s_n-s_m|=\left|\|x_n-(-y_n)\|-\|x_m-(-y_m)\|\right|\leq \|x_n-x_m\|+\|-y_n+y_m\|$$ and so since $(x_n),(y_n)$ are Cauchy, for any $\epsilon>0$ we have some $N\in\mathbb N$ such that $$n,m\geq N\implies \|x_n-x_m\|<\epsilon/2$$ and similarly we have some $N'\in\mathbb N$ such that such that $$n,m\geq N'\implies \|y_n-y_m\|<\epsilon/2$$ so $$n,m\geq \max\{N,N'\}\implies |s_n-s_m|\leq \|x_n-x_m\|+\|-y_n+y_m\|<\epsilon/2+\epsilon/2=\epsilon$$ thus $(s_n)$ is Cauchy.

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HINT:

$$\begin{align*} |s_n-s_m|&=|\,\|x_n+y_n\|-\|x_m+y_m\|\,|\\ &=|\,\|x_n-(-y_n)\|-\|x_m-(-y_m)\|\,| \end{align*}$$

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