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Graph of f:[0,1]->R compact <=> f continuous

  1. Jul 6, 2013 #1
    I know this proof is probably super easy but I'm really stuck. I don't want someone to solve it for me, I just want a hint.


    One way is trivial:

    suppose f continuous.
    [0,1] compact and the continuous image of a compact space is compact so f([0,1]) is compact


    Now the other direction...

    here's some basic facts i have to work with:

    1) R is metrizable
    2) R is hausdorf
    3) R is regular
    4) R is normal
    5) f([0,1]) is compact (by assumption) (so it follows that every open cover has a finite subcover)
    6) f([0,1]) is closed (follows from 5 because R is Hausdorf and compact subset of Hausdorf space is closed)
    7) R is NOT compact, but it's connected
    8) [0,1] is connected and compact (every open cover has finite subcover, and no separate of [0,1])


    I don't know, everything I try I Just get no where. Can someone give a hint on how/where to start? In showing f is continuous should I try the approach 1) pre-image under f of every open set U in f([0,1]) is open in [0,1] ? 2) pre-image under f of every closed set U in f([0,1]) is closed in [0,1] ? 3) maybe try a contradiction? Assume f is not continuous, which means there exists an open set U in f([0,1]) such that it's pre image is not open, etc.
     
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  3. Jul 6, 2013 #2

    micromass

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    So let ##G(f) = \{(x,f(x))~\vert~x\in [0,1]\}##. Maybe you can start by proving that ##[0,1]## and ##G(f)## are homeomorphic?
     
  4. Jul 6, 2013 #3

    WannabeNewton

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    dumbQuestion, your thread title says graph but in your main body it seems you are trying to show that if ##f([0,1])## is compact then ##f## is continuous but this is the image of ##f##, not the graph of ##f##, and it isn't true in general. Consider for example ##f:[0,1] \rightarrow \mathbb{R}## given by ##f(x) = 0, 0 \leq x\leq \frac{1}{2}## and ##f(x) = 1, \frac{1}{2} < x\leq 1##. The image of this is ##\{0,1\}## which is compact but the function isn't continuous. So are you sure you aren't supposed to actually look at the graph of ##f##?
     
  5. Jul 6, 2013 #4
    oh yes, I am supposed to look at the graph of f. I guess I was just mistaking the graph of f for the image of f. I guess I'm not seein gthe distinction between the graph and image.
     
  6. Jul 6, 2013 #5

    micromass

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    The graph of ##f## is by definition a subset of ##[0,1]\times \mathbb{R}##. While the image is a subset of ##\mathbb{R}##. They are two very different different things.
     
  7. Jul 6, 2013 #6
    oh i see. ok this will change things entirely. thank you so much for pointing this out
     
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