Can Continuity Guarantee a Minimum Value on an Interval?

[Tokipin]

Homework Statement

From Introduction to Topology by Bert Mendelson, Chapter 2.4, Exercise 8:

Let R be the real numbers and f: R -> R a continuous function. Suppose that for some number a \in R, f(a) > 0. Prove that there is a positive number k and a closed interval F = [a - \delta, a + \delta] for some \delta > 0 such that f(x) \geq k for x \in F.

Homework Equations

Neighborhood definition of continuity:

Let f:(X, d) -> (Y, d'). f is continuous at a point a \in X if and only if for each neighborhood M of f(a), f^{-1}(M) is a neighborhood of a.

The Attempt at a Solution

I think my proof is right, I just want to make sure because it's a bit more involved than the proofs I've done up to this point:

Let M be the closed ball about f(a) of radius f(a)/2. Because f is continuous, f^{-1}(M) is a neighborhood of a. By the definition of neighborhood, there is an open ball about a of radius \eta. All points in this ball map into M. Define a closed ball about a of radius \eta/2. This ball is the range [f-\delta,f + \delta] with \delta = \eta/2. It will map exclusively into M, which has a lower bound of f(a)/2. Set k = f(a)/2.

 

[rochfor1]

Youve got it! =)

 

[Tokipin]

Thanks for looking it over!