# Continuous mapping on cartesian product of metric spaces

Let $f:X\to X'$ and $g:Y\to Y'$ be continuous mappings. We'll define the decart product $Z:=X\times Y$ and metric $d_Z(z_1, z_2):=d_X(x_1,x_2)+d_Y(y_1,y_2)$ where $z_1=(x_1,y_1)$ and $z_2=(x_2,y_2)$.

We'll prove that $(f,g):X\times Y\to X'\times Y'$ is continuous.

Proof: $(f,g):X\times Y\to X'\times Y'$ is continuous at $(x_0,y_0)\in X\times Y$ if $\quad\forall \varepsilon>0$ $\exists \delta>0$ such that for any $(x,y)\in X\times Y$ and $d_Z((x,y),(x_0,y_0))<\delta$ implies that $$d_{Z'}((f(x),g(y)),(f(x_0),g(y_0)))<\varepsilon.$$

1. So $f$ is continuous at $x_0\in X$. For given $\varepsilon>0$ $\exists \delta_1>0$ such that for any $x\in X$ and $d_X(x,x_0)<\delta_1$ implies that $d_{X'}(f(x),f(x_0))<\varepsilon$.

2. So $g$ is continuous at $y_0\in Y$. For given $\varepsilon>0$ $\exists \delta_2>0$ such that for any $y\in Y$ and $d_Y(y,y_0)<\delta_2$ implies that $d_{Y'}(f(y),f(y_0))<\varepsilon$.

How from above conclusions I can get what I need?

• Does your product derive its name from René Descartes? If so, some re-spelling may be in order... – pjs36 Sep 20 '15 at 19:31
• @pjs36, Sorry please! I edited. – ZFR Sep 20 '15 at 19:33

• Can you explain why $\delta=2\min\{\delta_1, \delta_2\}$? – ZFR Sep 20 '15 at 19:41
• At first you wrote that we can take $\delta=2\min\{\delta_1,\delta_2\}$. But I think thas this delta does not work. If I am not true can you explain? – ZFR Sep 20 '15 at 20:04