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Proof on continuous functions

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    assume h: R->R is continuous on R and let K={x: h(x)=0}. Show that K is a closed set.

    2. Relevant equations

    3. The attempt at a solution

    since we know h is continuous and h(x)=0. therefore, we know there is a epsilon neighborhood such that x belongs to preimage f^(-1)({0}). I got stuck up to this step since h maps all x's in R to 0 so this could mean all members in R so i don't see it's a closed set at all. please help. i'm confused with this concept.
  2. jcsd
  3. Jun 1, 2008 #2
    Since {0} is closed in R. This means that [tex]h^{-1}{\{0\}}= \{x\in R:h(x)=0 \}[/tex]

    If the domain D is closed, then the inverse image of every closed set under h is also closed.
    Last edited: Jun 1, 2008
  4. Jun 1, 2008 #3


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    Or use the caracterisation of continuity by sequences. Let x_n be a sequence of element of K that converges in R to x. What can you say about h(x)?
  5. Jun 1, 2008 #4
    because {0} is closed in R and f is a function bijected R back to R. therefore, [tex]
    h^{-1}{\{0\}}= \{x\in R:h(x)=0 \}\supseteq {\{0\}} \supset R?
    Last edited: Jun 1, 2008
  6. Jun 2, 2008 #5


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    That makes no sense whatsoever. R isn't contained in {0} and there is no containment relation between h^(-1){0} and {0} either. Start really simple. Let h(x)=cos(x). What is h^(-1){0}? Is it closed? Just to make sure you actually understand what the problem is.
  7. Jun 2, 2008 #6
    sorry i mixed up the symbol it should be the reverse. and [tex] h^{-1}{\{0\}} \subset R [/tex].
    Last edited: Jun 2, 2008
  8. Jun 2, 2008 #7


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    That makes it a little better, but it's still not a proof. There's no 'bijection' here.
  9. Jun 2, 2008 #8


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    The very first time I was called upon to present a proof in graduate school, it involved "f-1(A)" for a set A. I assumed f had an inverse function and was very embarrassed when it was pointed out to me that the notation f-1(A) does NOT imply that. If f is, in fact, "one-to-one" then f-1({0})= {0} is a singleton set which is trivially closed. The problem is when f is NOT one-to-one and so is does not have an inverse function.

    It is certainly true that R\{0} is an open set. Can you use the fact that f-1({0})= R\ f-1(R\{0})?

    There is a standard topological characterization of continuous functions that makes that almost trivial but I don't know if you can use that.
  10. Jun 2, 2008 #9
    I like this one the best. If you consider a limit point of K, you can construct a sequence in K that converges to that point (by simply invoking the definition of a limit point), and thus show from the continuity of h that the limit point must then also be in K. If any limit point of K is in K, then K is closed by definition.
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