I want to prove that if a function, $f,$ is continuous on [a,b] such that $f(c)>0$, then there exists an open interval, $T,$ containing $c,$ such that $f(t)>0$ for all $t \in T$.
Here is my attempt:
Let $\epsilon >0.$ Since $f$ is a continuous function on $[a,b]$ there is a point $t=c$ such that there exists a $\delta>0.$ By the delta-epsilon definition of continuity, |$f(t)-f(c)|<\epsilon$ for all $t$ such that |$t-c|<\delta.$ This is equivalent to $c-\delta < t < c + \delta$ which gives that $t \in (c-\delta,c+\delta).$ Thus, the open interval, $T$ exists.
I'm not sure if this is correct or if there is an alternative solution to this that would be better. Any suggestions would be helpful.
So, you need to specify your $\epsilon$ and $\delta$ accordingly. Let us pick $\epsilon = \frac{f(c)}{2}$ (this is a quite standard way if you know the function value at one point and want to prove this type of results). Take, the corresponding $\delta$, whose existence is guaranteed by continuity, such that: $$ |t-c|<\delta \implies |f(t)-f(c)|<\epsilon $$ which implies that for every $t \in (c-\delta,c+\delta)$, $f(t)>\frac{f(c)}{2}>0$.