Jtdylktuy

If you have an equation of this kind: $u(x) × v(y)=0$
What would the set of solutions be?
Let s assume I have this one $u(x)^2 +v(y)^2 =0$
How different the set of solutions would be?

In order for $(x_0,y_0)$ to be a solution for $u(x) times v(y) =0$, you need either that $u(x_0)=0$ or that $v(y_0)=0.$ So if you call $X_0={x vert u(x)=0}$ and $Y_0={y vert v(y)=0}$, all the solutions to $u(x) times v(y) =0$ belong to the set $X_0 cup Y_0$.

For your second case (which is not a rearrangement of the first), you need both $u(x_0)=0$ and $v(y_0)=0.$ Therefore all solutions to $u(x)^2+v(y)^2=0$ belong to the set $X_0 cap Y_0$ which, as @saulspatz has noted, is a subset of the previous one $X_0 cap Y_0 subset X_0 cup Y_0$.

As $u(x)*v(y)=0$ that means whether $u(x)=0$ or $v(y)=0$ else, $u(x)$ and $v(y)$ both $=0$.and as you were able to make $u(x)^2+v(y)^2=0$(though you won't be able to do so) so it turns out both $u(x)=0$ and $v(y)=0$.So,now solve this two equation to get respective value of $x$ and $y$ for which $u(x)=0$ and $v(y)=0$.

If we consider these two equation as two different scenario then , for first equation there may be some value of x or y to make $u(x)$ or $v(y)=0$.but it can also be happen that $u(x)$ or $v(y) ne 0$.then there will be no solution for whether $x$ or $y$.but if the second equation is the case then there must be some solution of $x$ and $y$.so,we can say that ${x_{1_i},y_{1_i}}subset {x_{2_i},y_{2_i}}$.but the reverse is not true.

Would S for the first case be:

$S = $
${ (x_0, y); (x, y_0)/x,y in IR}$