# Proof on continuous functions

## Homework Statement

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

## 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.

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.

Last edited:
quasar987
Homework Helper
Gold Member
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)?

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

Last edited:
Dick
Homework Helper
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.

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

Last edited:
Dick
Homework Helper
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