# Thread: Prove that S is a closed set.

1. ## Prove that S is a closed set.

Let f:R→R be continuous and let S={x∈R:f(x)=0} be the set of all roots of f. Prove that S is a closed set.

How would you prove the above using the following: Let f : D→R be continuous at a point a∈D,and assume f(a)>0. Prove that there exists a δ > 0 such that f (x) > 0 for all x ∈ D ∩ (a − δ, a + δ).

I know I would need to use problem 2 to show that the compliment of S is open but Im not sure how.

2. Originally Posted by alice8675309
Let f:R→R be continuous and let S={x∈R:f(x)=0} be the set of all roots of f. Prove that S is a closed set.

How would you prove the above using the following: Let f : D→R be continuous at a point a∈D,and assume f(a)>0. Prove that there exists a δ > 0 such that f (x) > 0 for all x ∈ D ∩ (a − δ, a + δ).

I know I would need to use problem 2 to show that the compliment of S is open but Im not sure how.
Use the fact that $f^{-1}\left(\mathbb{R}-A\right)=\mathbb{R}-f^{-1}\left(A\right)$ for every $A\in\mathbb{A}$. Then using the basic idea of how you must have proven your second statement you can prove that the preimage of an open set is open and thus using that set-theoretic identity I mentioned the preimage of a closed set is closed.

3. Originally Posted by Drexel28
Use the fact that $f^{-1}\left(\mathbb{R}-A\right)=\mathbb{R}-f^{-1}\left(A\right)$ for every $A\in\mathbb{A}$. Then using the basic idea of how you must have proven your second statement you can prove that the preimage of an open set is open and thus using that set-theoretic identity I mentioned the preimage of a closed set is closed.
So wait, I'm confused on how the second statement is used to prove the pre image of an open set is open.

4. Originally Posted by alice8675309
So wait, I'm confused on how the second statement is used to prove the pre image of an open set is open.
You idea in the first post is best, I think.
Because $f$ continuous if $a\notin S$ then there is an open set $O_a$ such that $a\in O_a$ and if $y\in O_a$ then $|y|>0$.
But that means $S^c=\bigcup\limits_{a \in S^c } {O_a }$.
Thus showing $S^c$ is open.

5. Originally Posted by Plato
You idea in the first post is best, I think.
Because $f$ continuous if $a\notin S$ then there is an open set $O_a$ such that $a\in O_a$ and if $y\in O_a$ then $|y|>0$.
But that means $S^c=\bigcup\limits_{a \in S^c } {O_a }$.
Thus showing $S^c$ is open.
Ooo I get it now! Thanks so much you've been such a big help. For some reason I can never get the definitions of open and closed right and/ or use them correctly. Thanks a lot!