F is continuous then F is continuous in each variable separately

[DotKite]

Homework Statement

Let F: X x Y -> Z. We say F is continuous in each variable separately if for each $b \in Y$ the function h: X -> Z, h(x) = F(x,b), and for each $a \in X$, the function g: Y -> Z, g(y) = F(a,y) is continuous. Show that if F is continuous, then F is continuous in each variable separately.

Homework Equations

The Attempt at a Solution

So I noticed that you can define a function A: X -> X x Y and B: Y -> X x Y, and h = A ° F and
g = B ° F. If I can show that A and B are continuous then I am done. However I do not see how to do that. I have a theorem in my book that says,

Let the Topo space M be a subspace of the topo space N. Then the inclusion mapping i: M -> N is continuous.

Can I use this for A and B? Can I just say they are inclusion mappings?

 

[pasmith]

Homework Statement

Let F: X x Y -> Z. We say F is continuous in each variable separately if for each $b \in Y$ the function h: X -> Z, h(x) = F(x,b), and for each $a \in X$, the function g: Y -> Z, g(y) = F(a,y) is continuous. Show that if F is continuous, then F is continuous in each variable separately.

Homework Equations

The Attempt at a Solution

So I noticed that you can define a function A: X -> X x Y and B: Y -> X x Y, and h = A ° F and
g = B ° F.

You have your compositions backwards: if f: X \to Y and g: Y \to Z then
g \circ f : X \to Z : x \mapsto g(f(x)).

If I can show that A and B are continuous then I am done. However I do not see how to do that. I have a theorem in my book that says,

Let the Topo space M be a subspace of the topo space N. Then the inclusion mapping i: M -> N is continuous.

Can I use this for A and B? Can I just say they are inclusion mappings?

X is not a subset of X \times Y, so it's not that simple.

However, for each b \in Y there is an injection f : X \to X \times Y : x \mapsto (x,b). If you can show that f is continuous then you have that h = F \circ f : X \to Z : x \mapsto F(x,b) is continuous.

The basis of the topology on X \times Y is \{ U \times V : U is open in X and V is open in Y\}. If the pre-image of every basis set is open, then the pre-image of every open set is open (because an open set is an arbitrary union of basis sets, and the pre-image of an arbitrary union is the union of the pre-images).