Proof on continuous functions

In summary, we are given a continuous function h:R->R and we need to show that the set K={x:h(x)=0} is closed. We first note that since {0} is closed in R, the inverse image of {0} under h, written as h^{-1}{0}, is also closed. Using the characterization of continuity by sequences, we can choose a sequence x_n in K that converges to a limit point x. Since h is continuous, this implies that h(x_n) converges to h(x), and since each x_n is in K, we have that h(x_n)=0 for all n. Therefore, h(x)=0, and thus x is in K. This shows
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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.

Homework Equations





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.
 
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  • #2
iamjign said:
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
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
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
iamjign said:
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.
 
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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
iamjign said:
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
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
quasar987 said:
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.
 

What is a continuous function?

A continuous function is a mathematical function that has no abrupt changes or breaks in its graph. This means that the graph can be drawn without lifting the pencil from the paper.

What is the definition of a limit in continuous functions?

The limit of a function at a point is the value that the function approaches as the input value gets closer and closer to that point. In continuous functions, this means that the limit of the function at a point is equal to the value of the function at that point.

How can I prove that a function is continuous?

To prove that a function is continuous, you can use the three-part definition of continuity, which states that a function is continuous at a point if the limit at that point exists, the function is defined at that point, and the limit and the function value are equal.

What are the common methods of proving continuity?

The most common methods of proving continuity include the epsilon-delta method, the intermediate value theorem, and the sequential criterion for continuity. These methods involve using mathematical definitions and properties to show that the function is continuous at a given point or interval.

Why is continuity important in mathematics and science?

Continuity is important because it allows us to make meaningful predictions and observations about the behavior of functions. It also forms the basis for many other important mathematical concepts, such as differentiability and integrability, and is essential in fields such as physics and engineering.

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