Sign-preserving property of continuous functions Let $f$ be continuous at $c$ and suppose that $f(c) \neq 0$. Then there is an interval $(c - \delta, c + \delta)$ about $c$ in which $f$ has the same sign as $f(c)$.
The book (Apsotol calculus vol 1, page 143) has a constructive proof of this property however I wanted to try a proof by contradiction to see if it can be done that way as well.
For contradiction I assumed that in every interval $(c - \delta, c + \delta)$ about $c$ there exists a $x$ such that $f(x)$ has a sign different from $f(c)$. To make the argument specific assume $f(c) > 0$. Then by our assumption for every interval about $c$ there exists a $x$ in the interval such that $f(x) < 0$.
At this point I am stuck. What does it mean for a $x$ to exist in every interval about $c$ such that $f(x) < 0$ ? Does it mean that the function $f(x)$ is oscillating very rapidly? From here how to reach a contradiction?
Continuing: by IVT, there is some $a_{\delta} \in (c-\delta, c+\delta)$ such that $f(a_{\delta}) = 0$. This is true for every $\delta > 0$.
Note that $\lim_{\delta\to 0} a_{\delta} = c$, so by continuity of $f$, $\lim_{\delta \to 0} f(a_{\delta}) = f(c)$. Therefore, $f(c) = 0$. A contradiction.