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F is continuous then F is continuous in each variable separately

  1. Dec 9, 2013 #1
    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. jcsd
  3. Dec 9, 2013 #2

    pasmith

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    Homework Helper

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


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

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

    The basis of the topology on [itex]X \times Y[/itex] is [itex]\{ U \times V : U[/itex] is open in [itex]X[/itex] and [itex]V[/itex] is open in [itex]Y\}[/itex]. 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).
     
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