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F continuous on every compact subset; f cont. on the whole space?

  1. Dec 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Suppose that fk : X to Y are continuous and converge to f uniformly on every compact subset of the metric space X. Show that f is continuous.

    (fk is f sub k)

    2. Relevant equations
    Theorem from p. 150 of Rudin, 3rd ed:
    If {fn} is a sequence of continuous functions on E, and if fn converges unifromly on f on E, then f is continuous on E.


    3. The attempt at a solution
    Well, that's the thing. I can show that f is continuous on every compact subset of X. But how do I use that to show that f is continuous on X?

    I appreciate your help.
     
  2. jcsd
  3. Dec 16, 2007 #2

    quasar987

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    I think your problem is you're trying to show something but you don't have a clear idea what you're trying to prove. Let me ask you this.

    What does it mean to say f is continuous on X?
     
  4. Dec 16, 2007 #3
    Given epsilon positive, there exists positive delta s.t. d( f(x),f(y) ) < epsilon whenever d( x,y) < delta
     
  5. Dec 16, 2007 #4

    quasar987

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    Less formally, it means f is continuous at every point of X!
     
  6. Dec 16, 2007 #5
    right. but how do i prove that every point of X is part of a compact set? can i construct one? like, a closed ball around every point x would be compact, then i just have to choose the right radius. choose r s.t. B(x, r) is a subset of X, where B(x,r) is the closed ball of radius r centered at point x. okay, does that work?
     
  7. Dec 16, 2007 #6

    quasar987

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    The ball will always be in X whatever its radius. By definition, the ball of radius r centered on x is

    {y in X : d(x,y)<r}

    And yes, this is what I had in mind!

    But perhaps try to show that closed balls indeed are compact sets!
     
  8. Dec 16, 2007 #7
    That's good too. Also, how about, each x is a compact set (since for any open cover {Ga}, x is in some Ga1, so there is a finite subcover => x is compact), hence f is continuous at each x, hence f is continuous.

    does that work too?

    thank you!
     
    Last edited: Dec 16, 2007
  9. Dec 16, 2007 #8

    quasar987

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    I do not think this works.

    Because we can then only conclude that the function f:{x}-->Y is continuous, which is almost vacuously true and is not equivalent to saying that f:X-->Y is continuous at x.

    Cuz the result is.. "Let A be a subset of X. Then if fk:A-->Y (continuous) converges to f uniformly on A, then f:A-->Y is continuous on A." Arguing that fk converges to f on {x} allows us to conclude that f:{x}-->Y is continuous. This is not equivalent to saying that f:X-->Y is continuous at x because it does not imply that for any sequence xn converging to x, f(xn) converges to x, since the only sequence in {x} converging to x is the trivial sequence xn=x.
     
  10. Dec 18, 2007 #9
    okay. then what about points where such a ball consists of only a point?
     
  11. Dec 18, 2007 #10

    HallsofIvy

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    Singleton sets are trivially compact!
     
  12. Dec 19, 2007 #11
    right, that's what i thought. you could in fact do that for every point. but then quasar said that doesn't work...
    now i'm confused.
     
  13. Dec 19, 2007 #12

    quasar987

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    If it so happens that you take a ball around a point x and that point is the only point in the ball, then its okay. In the sense that you can conclude that the function is continuus there, because it means that the only sequence in X approaching x is the trivial sequence {x,x,x,x,....}.

    But if y is another point in the space X such that there are other sequences appraoching y other than {y,y,y,y,...}, then considering the compact set {y} only allows you to conclude that the function f:{y}-->Y is continuous, which is always true trivially since in {y}, the constant sequence {y,y,y,...} is the only sequence approaching y. But it does not mean that f:X-->Y is continuous at y because you know there are other sequences in X approaching y other that the constant sequnce {y,y,y,y,...}.

    Do you see a little better?
     
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