I am going through the second edition of Abbott's Understanding analysis. I am stuck on a problem regarding continuous functions.
Exercise:
Decide if the following claims are true or false, Providing either a short proof or counterexample to justify each conclusion. Assume throughout that $g$ is defined and continuous on all of $\mathbb{R}$.
(a) If $g(x) \geq 0$ for all $x<1$, then $g(1) \geq 0$ as well.
(b)if $g(r) = 0$ for all $r \in \mathbb{Q}$, then $g(x) = 0 $ for all $x \in \mathbb{R}$.
(c) if $g(x_{0}) >0 $ for a single point $x_0 \in \mathbb{R}$, then $g(x)$ is in fact strictly positive for uncountably many points.
My Attempt:
(a) True. Suppose $g(1) <0$. since $g$ is continuous on $\mathbb{R}$, it is continuous at $x=1$. Let $\epsilon = |g(1)|/2$. There exists $\delta >0$ such that $g(x) \in (g(1) – \epsilon, g(1) +\epsilon)$ for $x\in (1-\delta,1+\delta)$.
Here is where I am off track. I know this creates a contradition of the our assumption that $g(x) \geq 0$ for all $x<1$. But I don't know how to illustrate this.
(b) True. Let $x\in \mathbb{R}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a sequence $(r_n)$ in $\mathbb{Q}$ such that $r_n \rightarrow x$. By continuity of $g$, we have
$g(x) = g(\lim _{n \rightarrow \infty} r_n)= \lim_{n \rightarrow \infty} g(r_n) = 0$.
(c) Here I'd like to see this demonstrated and have the thought process explained. There are just many more conditions and I don't know how to start this or whether a short proof or a counterexample is more efficient/appropriate.
Best Answer
$c$ is true because we have a neigborhood around $x_0$ , $I=[x_0-\epsilon,x_0+\epsilon]$ , such that $f(x)>0$ for all $x\in I$.
And $I$ clearly contains uncountably many points.