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Quick topology question

  1. Dec 18, 2007 #1
    1. The problem statement, all variables and given/known data
    All right, so this appeared on my final. The intervals are in the reals:

    If f : [a, b] -> [c, d] , and the graph of f is closed, is f continuous?


    2. Relevant equations



    3. The attempt at a solution
    Well, my gut reaction is no, just because it seems like a fairly strong claim, but I couldn't really come up with a decent counterexample. On the other hand, I can see that the graph has to be compact. And I think that would imply that f([a, b]) has to be compact (if it wasn't you could just take any open cover of [a, b] and an open cover of f([a, b]) that doesn't have a finite subcover, then the Cartesian product would be an open cover without finite subcover, I think). But I'm not sure where I can go from there or if that's even useful information.
     
  2. jcsd
  3. Dec 19, 2007 #2

    Dick

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    If the sequence x_i->x in [a,b] then the sequence f(x_i) has cluster points in [c,d] because [c,d] is compact. Suppose it has two unequal ones y1 and y2. Then since the graph is closed y1 and y2 are both on the graph. Is this compatible with f being a single valued function?
     
  4. Dec 19, 2007 #3

    quasar987

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    But what guarentees that the limit of f(x_i) is f(x)?
     
  5. Dec 19, 2007 #4
    Bolzano-Weierstrass
     
  6. Dec 19, 2007 #5

    Dick

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    If f(x_i) has a limit y, then (x,y) is on the graph since the graph is closed. The graph is the set of all points (x,f(x)).
     
  7. Dec 19, 2007 #6

    quasar987

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    That's true. That was a good little problem!
     
  8. Dec 19, 2007 #7

    morphism

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    What about the converse: if f:[a,b]->[c,d] is continuous, then is its graph closed?
     
  9. Dec 19, 2007 #8

    quasar987

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    That one's easy. If (x_i, f(x_i)) is a converging sequence in the graph, then it means x_i -->x, for some x in [a,b], and by continuity, f(x_i)-->f(x). Since (x,f(x)) is in the graph, it's closed.
     
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