Proof on continuous functions

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.

 

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)?

 

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.

 

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

 

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.

 

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.