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If f,g cont. , then g o f cont.?

  1. Oct 9, 2008 #1
    If f,g cont. , then "g o f" cont.?

    1. The problem statement, all variables and given/known data
    Let X,Y,Z be spaces and f:X-->Y and g:Y-->Z be functions.
    If f,g are continuous,
    then so is g o f.


    2. Relevant equations



    3. The attempt at a solution

    This is my proof so far. I would be appreciative if someone could point out the weak points.

    "Proof". Let f,g be continuous. Show "g o f" continuous. Show g(f(x)) continuous.

    Since f is continuous, and since f:X-->Y, we know that x is an element of W such that f[W][tex]\subseteq[/tex]V for V open, V[tex]\subseteq[/tex]Y.

    Since g is continuous, and since g:Y-->Z, we know that f(x) is an element of V such that g[V][tex]\subseteq[/tex]M, for M open, M[tex]\subseteq[/tex]Z.

    Therefore, g[ f(x) ] [tex]\in[/tex] M [tex]\subseteq[/tex] Z.

    Thus, "g o f" is continuous.
     
  2. jcsd
  3. Oct 9, 2008 #2
    Re: If f,g cont. , then "g o f" cont.?

    Let me try to reiterate it. Say we have a neighborhood of f(x), V. Then we can find a neighborhood of x, U in X such that f(U) is in V. In the same way, if we have a neighborhood of g(f(x)) W in Z, we can find a neighborhood of f(x) T in Y, such that g(T) is in W. So, we should be able to find a neighborhood of x, S in X such that g(f(S)) is a subset of W.
     
    Last edited: Oct 9, 2008
  4. Oct 9, 2008 #3
    Re: If f,g cont. , then "g o f" cont.?

    It sounds alot better your way. I understand it more now.
     
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