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!