# Proof check for completeness

I'd like to see if the proof I have is adequate.

Statement. Let $X$ and $Y$ be Banach space, the product $X\times Y$ is a vector space under coordinate operations with norm $$\|(x,y)\| = \|x\|_X +\|y\|_Y.$$ Show that we have a Banach space with the defined norm.

Proof. Since $X$ and $Y$ are both Banach spaces, then $\forall \, \varepsilon>0 \ \exists N$ such that if $n>N$ then $$\|x_n -x\|_X < \frac{\varepsilon}{4}\quad\text{and}\quad\|y_n-y\|_Y < \frac{\varepsilon}{4}.$$

Hence if we start of with a Cauchy sequence $\{(x_n,y_n)\}$ we have:

$\begin{eqnarray*} \|(x_n,y_n)-(x_m,y_m)\| &=& \|(x_n-x_m,y_n-y_m\| \\ &=& \|x_n-x_m\|_X + \|y_n-y_m\|_Y \\ &\leq& \|x_n-x\|_X + \|x-x_m\|_X + \|y_n-y\|_Y + \|y-y_m\|_Y \\ &<& \frac{\varepsilon}{4} + \frac{\varepsilon}{4}+ \frac{\varepsilon}{4}+ \frac{\varepsilon}{4} = \varepsilon \end{eqnarray*}$

So we see that our Cauchy sequence is convergent.

• In addition to what @Yiorgos stated To see that $\lim_{n\rightarrow \infty} \{(x_n, y_n)\} = (x, y)$ we have $\begin{eqnarray*} ||(x_n,y_n)-(x,y)|| &=& ||(x_n-x,y_n-y)|| \\ &=& ||x_n-x||_X + ||y_n-y||_Y \\ &<& \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{eqnarray*}$ – nonameswereavailable Feb 20 '14 at 22:52

In your proof, you should start with a Cauchy sequence $\{(x_n,y_n)\}\subset X\times Y$ and show that this sequence is convergent in $X\times Y$.
First observe that if $\{(x_n,y_n)\}$ is a Cauchy sequence in $X\times Y$, then both $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences in $X$ and $Y$ respectively, since $$\|x_m-x_n\|+\|y_m-y_n\|=\|(x_m,y_m)-(x_n,y_n)\|<\varepsilon.$$
Now, as $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences in $X$ and $Y$ respectively, and $X$ and $Y$ are Banach spaces, then they converge say to $x$ and $y$, respectively.
Finally, show that $\{(x_n,y_n)\}$ converges to $(x,y)$, and thus every Cauchy sequence in $X\times Y$ is convergent.