Let $p$ be a point in $X$ at which $f(p) \gt 0$ Show there is some positive number $r$ such that $f(q) \gt 0$ for all q in $B_r(p)$

Question

Let $X$ be a metric space and let the mapping $f : X\rightarrow R$ be continuous. Let $p$ in $X$ s.t. $f(p) \gt 0$. By epsilon-delta definition of continuity, Show there is some positive number $r$ such that $f(q) \gt 0$ for all $q$ in $B_r(p)$

What I have right now is for $\epsilon \gt 0$ there exists $\delta \gt 0$ s.t. $d(p,q) \lt \delta \Rightarrow d(p,q) \lt \epsilon$. Then we can get $f(B_\delta(p)) \subseteq B_\epsilon(f(p))$. I'm assuming $r = \delta$ is my answer but I don't know how to show that $q \in B_\delta(p)$ implies that f(q) > 0. Any suggestions?

Answer 1

Take $\epsilon = f(p)/2$ and use the continuity of $f$ at $p$ to obtain the radius $r = \delta$ of the open ball $B_\delta(p)$ such that $f(B_\delta(p)) \subseteq B_\epsilon(f(p))$, meaning that $d(p,q) \lt \delta \implies |f(p)-f(q)| \lt \epsilon = f(p)/2$. In this case, $$f(q) - f(p) > -\epsilon = -f(p)/2, \\ f(q) > f(p) - f(p)/2 = f(p)/2 > 0.$$

Answer 2

Or proof without $\varepsilon-\delta$ by contradiction:

Assume $$\forall r > 0 \exists q_r \in B_r(p): \, f(q_r) \leq 0$$ So, you can construct a sequence: $$\mbox{For }r_n = \frac{1}{n}\; (n \in \mathbb{N})\mbox{ choose } p_n \in B_{\frac{1}{n}}(p):\; f(p_n) \leq 0 \Rightarrow p_n \stackrel{n\rightarrow \infty}{\longrightarrow}p $$$$\stackrel{f \mbox{ continuous}}{\Rightarrow} f(p_n)\stackrel{n\rightarrow \infty}{\longrightarrow}f(p)\leq 0 \Rightarrow \mbox{ contradiction to } f(p) > 0$$