Hi,

i just learned the notions of jointly and separately cont. functions.

A function f: X x Y -> A is called

1) jointly cont <=> f is cont. w.r.t. the product topology

2) separ. cont <=> f$\displaystyle _x : Y->A, f_x (y)=(x,y)$ and $\displaystyle f_y$ are cont.

So i tryed to show 1) => 2).

But all my attempts failed.

Can you help me?

One of my attemts:

Let $\displaystyle f: X \times Y->A$ be cont. then $\displaystyle \forall U \subset A$ we get an open set $\displaystyle X' \times Y':=f^{-1} (U) \subset X \times Y$ ., whereas X' and Y' are open in X, Y respectively.

If we now consider the map $\displaystyle f_x : Y->A $ and take some U \subset A.

Now we have to proof, $\displaystyle (f_x) ^{-1}(U)=\{y \in Y : (x,y) \in A\}$ is open in Y. But why is this correct in all top. spaces?

Regards