Question : Let $f: \Bbb R \rightarrow \Bbb R$ be continuous. Suppose $f(c) >0$. Show that there exists an $\alpha>0$ such that for all $x \in (c-\alpha, c+\alpha)$ we have $f(x)>0$.
Attempt: By definition, there exists $\delta>0$ such that if $|x-c|<\delta$, then $|f(x)-f(c)|<\varepsilon $ for $\varepsilon >0$.Since $f(c)$ is larger than $0$, $f(x)$ must be larger than $0$, otherwise this condition
"$|f(x)-f(c)|<\varepsilon$" cannot be satisfied for all $\varepsilon>0$.
Could you tell me whether the proof is correct or not?
Thank you in advance!