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Possible Duplicate:
Sum of Cauchy Sequences Cauchy?

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

Show that if $(x_{n})_{n}$ and $(y_{n})_{n}$ are Cauchy sequences in $X$, then the sequence $(x_{n}+y_{n})_{n}$ is also Cauchy in $X$.

I have used the definitions:
If $(x_{n})_{n}$ is Cauchy, $\forall\epsilon>0:\exists N\in\mathbb{N}:n,m\ge N\implies||x_{n}-x_{m}||<\frac{\epsilon}{2}$
If $(y_{n})_{n}$ is Cauchy, $\forall\epsilon>0:\exists M\in\mathbb{N}:n,m\ge M\implies||y_{n}-y_{m}||<\frac{\epsilon}{2}$

To come up with $||x_{n}+x_{m}|| + ||y_{n}+y_{m}|| \le \epsilon$

How can I rephrase the left hand side of the inequality to come up with $||(x_{n}+y_{n}) - (x_{m}+y_{m})||$ to show Cauchyness?

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marked as duplicate by Kannappan Sampath, mixedmath, Zev Chonoles Mar 1 '12 at 15:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Triangle inequality. – Asaf Karagila Feb 29 '12 at 15:50
I tried the triangle inequality, how're you supposed to go backwards with it to get it into the form desired? – dplanet Feb 29 '12 at 15:51
Don't you mean $\|x_n - x_m\| + \|y_n - y_m\| \leq \epsilon$? Then do what Ragib suggested below. – Neal Feb 29 '12 at 16:01
I am voting to close this as Duplicate! – user21436 Mar 1 '12 at 7:21
up vote 3 down vote accepted

You were almost there! Rewrite $ \| (x_n + y_n ) - (x_m+ y_m) \| = \| (x_n - x_m) + (y_n - y_m) \| $ and apply the triangle inequality.

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