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Proof on continuous functions

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Homework Statement



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

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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.
 

Answers and Replies

  • #2
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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.
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.
 
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  • #3
quasar987
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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)?
 
  • #4
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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?
[/tex]
 
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  • #5
Dick
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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?
[/tex]
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.
 
  • #6
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sorry i mixed up the symbol it should be the reverse. and [tex] h^{-1}{\{0\}} \subset R [/tex].
 
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  • #7
Dick
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sorry i mixed up the symbol it should be the reverse. and [tex] h^{-1}{\{0\}} \subset R [/tex].
That makes it a little better, but it's still not a proof. There's no 'bijection' here.
 
  • #8
HallsofIvy
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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.
 
  • #9
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)?
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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