Is K a Closed Set for Continuous Functions?

[iamjign]

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.

 

[konthelion]

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 $$h^{-1}{\{0\}}= \{x\in R:h(x)=0 \}$$

If the domain D is closed, then the inverse image of every closed set under h is also closed.

 

[quasar987]

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)?

 

[iamjign]

because {0} is closed in R and f is a function bijected R back to R. therefore, $$h^{-1}{\{0\}}= \{x\in R:h(x)=0 \}\supseteq {\{0\}} \supset R?$$

 

[Dick]

because {0} is closed in R and f is a function bijected R back to R. therefore, $$h^{-1}{\{0\}}= \{x\in R:h(x)=0 \}\supseteq {\{0\}} \supset R?$$

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.

 

[iamjign]

sorry i mixed up the symbol it should be the reverse. and $$h^{-1}{\{0\}} \subset R$$.

 

[Dick]

sorry i mixed up the symbol it should be the reverse. and $$h^{-1}{\{0\}} \subset R$$.

That makes it a little better, but it's still not a proof. There's no 'bijection' here.

 

[HallsofIvy]

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.

 

[DavidWhitbeck]

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.