If f is continuous at a and f(a) > 0, then there is δ > 0 such that f(x) > 0 for all x ∈ dom(f) with |x − a| < δ I'm having trouble proving this statement especially because I am not fully convinced it is true.
"If f is continuous at a and f(a) > 0, then there is δ > 0 such that
f(x) > 0 for all x ∈ dom(f) with |x − a| < δ"
I know that given that f is continuous at a, for any >0, there exists  such that for any |−|<, |()−()|< holds, i.e., there exists an open interval =(−,+), such that for any ∈, |()−()|< holds.
Do I need to show that f(x) is within the interval for a carefully chosen δ? I am not sure where to go from here.
Thanks!

Yes. As $f(a) > 0$ there exists $\epsilon > 0$ such that $f(a) - \epsilon> 0$ (Take, for instance, $\epsilon = \frac{f(a)}{2}$). From the definition of continuity, for such $\epsilon > 0$ there exists $\delta > 0$ such that for every $x \in Dom(f)$, $|x - a| < \delta$ we have $|f(a) - f(x)| < \epsilon$, which means that $0 < f(a) - \epsilon < f(x) < f(a) + \epsilon$. This is a particular case of the result which says that for every open set $A \subset \mathbb{R}$ we have that $f^{-1}(A)$ is an open set.
