I was solving Stephen Abbot exercises and I came across this one I'm struggling with:
Assume $g$ is differentiable at point $c \in (a,b)$. If $g'(c) \neq 0$ show that there exists $\delta $ neighborhood around $c \in (a,b)$ for which $g(x)\neq g(c), \forall x$ in that neighborhood .
Although I think I understand intuition behind this, I still don't now how to approach this problem. I think I have to prove it by contradiction, assuming that for all $\delta$ neighborhoods around $c$, $g(x) = g(c)$ for at least one $x$, but I don't know where to go starting from here. I think, at some point I'll have to end up stating that in that case $g'(c)$ has to be equal to zero, which would be the desired contradiction.
I also tried to build sequences in $(a,b)$ that converge to $c$, but apparently that didn't help at all.
Your approach works indeed: suppose for sake of contradiction that for all $\delta$-neighbourhoods of $c$, there is an $x \ne c$ in said neighbourhood such that $g(x) = g(c)$. In particular, we can find for all integers $n$ big enough an $x_n \in (c - 1/n, c + 1/n)$ such that $x_n \ne c$ and $g(x_n) = g(c)$. Here, big enough means such that $(c - 1/n, c + 1/n) \subseteq (a, b)$. Now, what can you say about the sequence $(x_n)_n$?