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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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    $\begingroup$ Triangle inequality. $\endgroup$
    – Asaf Karagila
    Feb 29, 2012 at 15:50
  • $\begingroup$ I tried the triangle inequality, how're you supposed to go backwards with it to get it into the form desired? $\endgroup$
    – dplanet
    Feb 29, 2012 at 15:51
  • $\begingroup$ Don't you mean $\|x_n - x_m\| + \|y_n - y_m\| \leq \epsilon$? Then do what Ragib suggested below. $\endgroup$
    – Neal
    Feb 29, 2012 at 16:01
  • $\begingroup$ I am voting to close this as Duplicate! $\endgroup$
    – user21436
    Mar 1, 2012 at 7:21

1 Answer 1

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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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