Analysis of Continuous functions

[Locoism]

Homework Statement

Let f : R → R be continuous on R and assume that P = {x ∈ R : f(x) > 0} is non-empty. Prove that for any x0 ∈ P there exists a neighborhood Vδ(x0) ⊆ P.

Homework Equations

The Attempt at a Solution

If you choose some x, y ∈ P, since f(x) is continuous then |f(x) - f(y)| < ε for some ε>0
then |x-y|<δ for some δ(ε)

I don't really know where I'm going with this, but I know it has something to do with the question...
Can someone point me in the right direction?

 

[LCKurtz]

Homework Statement

Let f : R → R be continuous on R and assume that P = {x ∈ R : f(x) > 0} is non-empty. Prove that for any x0 ∈ P there exists a neighborhood Vδ(x0) ⊆ P.

Homework Equations

The Attempt at a Solution

If you choose some x, y ∈ P, since f(x) is continuous then |f(x) - f(y)| < ε for some ε>0
then |x-y|<δ for some δ(ε)

I don't really know where I'm going with this, but I know it has something to do with the question...
Can someone point me in the right direction?

You have to start with a correct statement of what continuity at x means. Finish this sentence correctly:

The statement that f(x) is continuous at x means ...

 

[Locoism]

that for any ε>0 there exists δ(ε) such that for any y, if |x-y|<δ then |f(x)-f(y)|<ε.
I don't see how this is different from what I put down?

 

[Locoism]

I'm finding it hard just to put it down into mathematical terms. I can prove it by sketching a drawing, but having trouble translating that...

 

[LCKurtz]

that for any ε>0 there exists δ(ε) such that for any y, if |x-y|<δ then |f(x)-f(y)|<ε.
I don't see how this is different from what I put down?

That may be why you are having difficulty with this type of problem. But leaving that aside for now, you have a value x₀ where f(x₀) > 0. The above statement with x = x₀ says you can get f(y) close to f(x₀). If you can do that is there some way you can guarantee f(y) is positive?