Show that if a continuous function is positive at a point it is positive on an interval around that point 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!
 A: The definition of continuity is that $f$ is continuous at $x$ if
$$ (\forall \epsilon >0)(\exists \delta >0)\ \text{such that}\ \lvert y - x \rvert < \delta \implies \lvert f(y) - f(x)\rvert < \epsilon$$
The definition you seem to be using is that $f$ is continuous at $x$ if
$$ (\exists \delta >0)(\forall \epsilon > 0)\ \text{such that}\ \lvert y - x \rvert < \delta \implies \lvert f(y) - f(x)\rvert < \epsilon$$
These are two very different definitions. This second definition actually says that there exists $\delta > 0$ such that $\lvert y - x \rvert < \delta$ then $f(y) = f(x)$, i.e. that the function is constant in a little neighbourhood around $x$. 
The way to solve this question is to pick a specific value of $\epsilon$ then to use continuity to find a $\delta$ which works. This $\delta$ will be the $\alpha$ in the question. What value of $\epsilon$ should you use?
A: $$\text{Def: we say $f$ is continuous at $c$ if:}\\(\forall \varepsilon >0)(\exists \delta >0)(\forall x)\left(\lvert x - c \rvert < \delta \implies \lvert f(x) - f(c)\rvert < \varepsilon\right)$$
Choose $\varepsilon=f(c)$, because it is given that $f$ is continuous I know that exists $\delta$ such that if $x\in(c-\delta,c+\delta)$ we have $-f(c)< f(x)-f(c)<f(c)\implies0<f(x)<2f(c)\implies f(x)\text{ positive}$

What you did is wrong because you search what $\delta$ is good for all $\varepsilon$, and it should be the other way around, for all $\varepsilon$ there exists $\delta$
If something is unclear ask, the definition of continuity can be confusing at the beginning
