“$(x_n, y_n)$ is weakly convergent to $(x,y)$” implies “$(x_n)$ is weakly converging to $x$”?

Let $$H_1$$, $$H_2$$ and $$H_1 \times H_2$$ be three Hilbert spaces, let the sequence $$(x_n, y_n)$$ be weakly convergent to $$(x,y)$$ in $$H_1 \times H_2$$. Then, when do we have " the sequence $$(x_n)$$ is weakly converging to $$x$$ and the sequence $$(y_n)$$ is weakly converging to $$y$$"?

Recall that the inner product on $$H_1\times H_2$$ is given by $$\langle(x_1,y_1),(x_2,y_2)\rangle=\langle x_1,x_2\rangle+\langle y_1,y_2\rangle.$$
Now to the problem at hand: If $$\{(x_n,y_n)\}$$ converges weakly to $$(x,y)$$, then for all $$x'\in H_1$$, we have $$\langle x_n,x'\rangle=\langle x_n,x'\rangle+\langle y_n,0\rangle\to \langle x,x'\rangle+\langle y,0\rangle=\langle x,x'\rangle$$ as $$n\to\infty$$. Thus $$\{x_n\}$$ converges weakly to $$x$$, and similarly we see that $$\{y_n\}$$ converges weakly to $$y$$.