# F is continuous then F is continuous in each variable separately

1. Dec 9, 2013

### DotKite

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

2. Dec 9, 2013

### pasmith

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

$X$ 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).