Let $f:\mathbb R^2\to \mathbb R^2$, $f(x,y)=(g(x,y),h(x,y))$ where $g,h :\mathbb R^2 \to \mathbb R$. $f$ is continuous at $P_0$ iff both $g,h$ are continuos at $P_0$
My attempt:
$\Rightarrow$:
Since $f$ is continuous at $p_0=(x_0,y_0)$ then $\forall \epsilon>0,\exists \delta>0$, s.t. If $\delta>||p-p_0||=\sqrt{(x-x_0)^2+(y-y_0)^2}$ $\Rightarrow$$\epsilon>||f(p)-f(p_0)||=\sqrt{(h(x,y)-h(x_0,y_0))^2+(g(x,y)-g(x_0,y_0))^2}$
And we know $\sqrt{(h(x,y)-h(x_0,y_0))^2+(g(x,y)-g(x_0,y_0))^2}\ge \sqrt{(g(x,y)-g(x_0,y_0))^2}=||g(p)-g(p_0)||$
and
$\sqrt{(h(x,y)-h(x_0,y_0))^2+(g(x,y)-g(x_0,y_0))^2}\ge \sqrt{(h(x,y)-h(x_0,y_0))^2}=||h(p)-h(p_0)||$
Therefore, for any given $\epsilon>0$, choosing the same $\delta>0$ we got, If $\delta>||p-p_0||$ then $\epsilon>||h(p)-h(p_0)||$ and $\epsilon>||g(p)-g(p_0)||$
$\Leftarrow$:
Now use the following:
$\sqrt{(h(x,y)-h(x_0,y_0))^2+(g(x,y)-g(x_0,y_0))^2}\le \sqrt{(g(x,y)-g(x_0,y_0))^2}+\sqrt{(h(x,y)-h(x_0,y_0))^2}$
Now $g,h$ are continuous at $p_0$ so $\forall \epsilon/2 \;,\exists\delta>0$ st $\delta>||p-p_0||$ then $\epsilon/2>||g(p)-g(p_0)||$ and $\epsilon/2>||h(p)-h(p_0)||$
Therefore, $||f(p)-f(p_0)||=\sqrt{(h(x,y)-h(x_0,y_0))^2+(g(x,y)-g(x_0,y_0))^2}<\epsilon$
Question: Is this valid proof? I am not sure, thank you in advanced.