Topology: Continuous f such that f(u)>0 , prove ball around u exists such that

[Bosley]

Homework Statement

Let O be an open subset of R^n and suppose f: O --> R is continuous. Suppose that u is a point in O at which f(u) > 0. Prove that there exists an open ball B centered at u such that f(v) > 1/2*f(u) for all v in B.

Homework Equations

f continuous means that for any {u_k} in O that converges to some point u, f(u_k) converges to f(u).

The Attempt at a Solution

Consider the open ball B_r(u) with r=(1/2)*f(u). Suppose v is in B_r(u). Then ||v - u || < (1/2)*f(u). Also,
|(1/2)*f(u)| <= |(1/2)*f(u) - f(v)| + |f(v)| so,
|f(v)| >= |.5*f(u)| – |.5*f(u) - f(v)|

The above statements are true but they're not getting me anywhere. I'd appreciate any help you can offer.

 

[losiu99]

Let's use Cauchy definition of continuity: since $$f(u)>0$$, then $$(1/2)f(u)>0$$ as well. There exist a ball $$B$$ around $$u$$ such that, for all $$v\in B$$ we have $$|f(v)-f(u)|<(1/2)f(u)$$.
But this implies in particular
$$f(u)-f(v)<(1/2)f(u)$$
and so $$f(v)>(1/2)f(u)$$.