Yes, it's indeed the idea.
There are some typos in your post. , not and we should have (we know that is open, but the open sets are not always product of open sets). If we denote by , we have and we are done since we know what is.
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 and are cont.
So i tryed to show 1) => 2).
But all my attempts failed.
Can you help me?
One of my attemts:
Let be cont. then we get an open set ., whereas X' and Y' are open in X, Y respectively.
If we now consider the map and take some U \subset A.
Now we have to proof, is open in Y. But why is this correct in all top. spaces?
Regards
Yes, it's indeed the idea.
There are some typos in your post. , not and we should have (we know that is open, but the open sets are not always product of open sets). If we denote by , we have and we are done since we know what is.
It's wasn't obvious for you because it was false.
To show the result, let and an open set which contains . By definition of the continuity and the product topology, we can find and two open sets with . Now, for we have hence .