Show that $R×R$ is a Hilbert Space. Let $R$ a Hilbert space. Show that $R×R$ is a Hilbert Space.
Clearly we can define the inner product on $R×R$ as the sum of the product of coordinates of points. How to show that it is complete metric space. Please guide me.
 A: Let $(x_n, y_n)_{n \in \mathbb N}$ a Cauchy sequence in $R \times R$, then there exists for each $\epsilon > 0$ a $N \in \mathbb N$ with $\Vert (x_n, y_n) - (x_m, y_m) \Vert_{R \times R} < \epsilon$ for all $n, m \geq N$. We need to show that this sequence converges in $R \times R$. As you noticed we have $\Vert (x_n, y_n) - (x_m, y_m) \Vert_{R \times R} = \Vert x_n - x_m \Vert_{R} + \Vert y_n - y_m \Vert_{R}$ and thus $(x_n)_{n \in \mathbb N}$ and $(y_n)_{n \in \mathbb N}$ are Cauchy sequences in $R$ and thus converge, say $x := \lim_{n \in \mathbb N} x_n$ and $y := \lim_{n \in \mathbb N} y_n$. Therefore we have
$$\Vert (x_n, y_n) - (x, y) \Vert_{R \times R} = \Vert x_n - x \Vert_{R} + \Vert y_n - y \Vert_{R}  < \epsilon  + \epsilon = 2 \epsilon$$
for $n$ small enough. So we showed that $(x_n, y_n)_{n \in \mathbb N}$ converges and since the sequence was arbitrarily choosen, it follows that $R \times R$ is complete and hence a Hilbert space. I hope that helps you :)
A: I suppose that $R = \mathbb R$. The norm induced by the inner product is $||(x,y)||=\sqrt{x^2+y^2}$.
We have 
(*) $|x| \le ||(x,y)|| \le |x|+|y|$ and $|y| \le ||(x,y)|| \le |x|+|y|$
It is your turn to derive from (*):
$(x_n,y_n))$ is a Cauchy sequence in $\mathbb R \times \mathbb R \iff$ $(x_n)$ and $(y_n)$ are Cauchy sequences in $ \mathbb R$.
