Convergence in inner product space Let $X$ be an inner product space. If $x_n\to x$, $y_n\to y$ (in norm), then $(x_n,y_n)\to(x,y)$ (in modulus).
 A: Add and subtract $(x,y_n)$ and use Cauchy's and triangle inequality many times.
$$\begin{align}|(x,y)-(x_n,y_n)|&=|(x,y)-(x,y_n)+(x,y_n)-(x_n,y_n)|\\&\leq |(x,y-y_n)|+|(x-x_n,y_n)|\\&\leq ||x||\cdot||y-y_n||+||x-x_n||\cdot||y_n||\\&\leq||x||\cdot||y-y_n||+||x-x_n||\cdot(||y||+\epsilon)\end{align}$$
for $n$ large such that $||y_n||-||y||\leq ||y_n-y||<\epsilon$.
A: I'm not sure what you mean by "modulus" but other definition of the norm on $X \times Y$ could probably be done with very similar proofs.
Let $\|\cdot\|_X,\|\cdot\|_Y$ be norms on $X$ and $Y$ and $\|\cdot\|_{X\times Y}:X \times Y \to \Bbb R: (u,v)\mapsto \sqrt{\|u\|_X^2+\|v\|^2_2}$. Let us suppose that $(x_n)_{n\in\Bbb N} \subset X$ and $(y_n)_{n\in\Bbb N} \subset Y$ are sequences so that 
$$\lim_{n\to\infty}\|x_n-x\|_X =0 \quad \text{ and } \quad \lim_{n\to\infty}\|y_n-y\|_Y =0.$$
Let $\epsilon > 0$, there exists $N_1,N_2>0$ such that $$\|x_n-x\|_X < \frac{\epsilon}{\sqrt 2}\qquad \forall n > N_1$$ and $$\|y_n-y\|_Y < \frac{\epsilon}{\sqrt 2}\qquad \forall n > N_2.$$ Then for every $n > \max\{N_1,N_2\}$ we get
$$\|(x_n,y_n)-(x,y)\|_{X\times Y}= \sqrt{\|x-x_n\|_X^2+\|y_n-y\|_2}\leq \sqrt{\left(\frac{\epsilon}{\sqrt 2}\right)^2+\left(\frac{\epsilon}{\sqrt 2}\right)^2} = \epsilon.$$
That is $(x_n,y_n)\to (x,y)$ in the norm $\|\cdot\|_{X \times Y}$
