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Math Help - how to prove that x: 0<=f(x)<=1 is compact

  1. #1
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    how to prove that x: 0<=f(x)<=1 is compact

    Suppose f: D-R is continuous with D compact. prove that x: 0<=f(x)<=1 is compact.

    what i got so far is that since D is compact, then D is closed and bounded. and f is uniformly continuous and f(D) is compact. but i dont how to show that x: 0<=f(x)<=1 is compact.

    Thanks!
    Last edited by Plato; April 4th 2010 at 03:18 PM.
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  2. #2
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    Can you show that \{x\in D:f(x)\le 1\} is a closed set?
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  3. #3
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    how? can we say that because it is a uniformly continuous function?
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    So if 0 \leq f(x_n) \leq 1 with  x_n \rightarrow x , with f continuous, then would we have  0 \leq f(x) \leq 1 ?
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    Quote Originally Posted by shuxue View Post
    Suppose f: D-R is continuous with D compact. prove that x: 0<=f(x)<=1 is compact.

    what i got so far is that since D is compact, then D is closed and bounded. and f is uniformly continuous and f(D) is compact. but i dont how to show that x: 0<=f(x)<=1 is compact.

    Thanks!
    Quote Originally Posted by Plato View Post
    Can you show that \{x\in D:f(x)\le 1\} is a closed set?
    I think the point Plato is trying to make is that if \to \mathbb{R}" alt="f\to \mathbb{R}" /> is continuous we have that f^{-1}\left([0,1]\right)\subseteq D is closed. But, a closed subspace of a compact space is compact.
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  6. #6
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    [0,1] is closed in R. Inverse image of closed set under continuous map is closed. Closed subset of a compact space is compact.

    >then D is closed and bounded
    That's already false since you don't have any relationship between D and R, I mean whether D is a subset of R or not. The Bolzano–Weierstrass theorem only applies when D is a subset of R^n
    Last edited by FancyMouse; April 4th 2010 at 10:04 PM. Reason: add some comments
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I think the point Plato is trying to make is that if \to \mathbb{R}" alt="f\to \mathbb{R}" /> is continuous we have that f^{-1}\left([0,1]\right)\subseteq D is closed. But, a closed subspace of a compact space is compact.
    Quote Originally Posted by FancyMouse View Post
    [0,1] is closed in R. Inverse image of closed set under continuous map is closed. Closed subset of a compact space is compact.
    Startling resemblance.
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    Quote Originally Posted by Drexel28 View Post
    Startling resemblance.
    Oops. I didn't notice your reply when I was writing.
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by FancyMouse View Post
    That's already false since you don't have any relationship between D and R, I mean whether D is a subset of R or not. The Bolzano–Weierstrass theorem only applies when D is a subset of R^n
    I don't follow. In any topological space we have that the inverse image of a closed set under a continuous map is closed, and that the closed subspace of a compact space is compact. There is no need for the B.W.P. here
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  10. #10
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    Quote Originally Posted by Drexel28 View Post
    I don't follow. In any topological space we have that the inverse image of a closed set under a continuous map is closed, and that the closed subspace of a compact space is compact. There is no need for the B.W.P. here
    I'm commenting on shuxue's statement that "what i got so far is that since D is compact, then D is closed and bounded"
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    Does anyone know/have the actual proof to this problem? It looks interesting and I was wondering if I could see the proof. (I'd attempt it but i don't want to start mixing my classes. Therefore i kind of just want to compare and contrast)
    Thanks
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alice8675309 View Post
    Does anyone know/have the actual proof to this problem? It looks interesting and I was wondering if I could see the proof. (I'd attempt it but i don't want to start mixing my classes. Therefore i kind of just want to compare and contrast)
    Thanks
    It's exactly what I said. Since f is continuous we have that f^{-1}([0,1])=\left\{x:0\leqslant f(x)\leqslant 1\right\} is a closed subspace of D. But, D is compact and so any closed subspace is automatically compact.
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