In my complex analysis textbook the following theorem is stated:
Theorem. Given a function $f$ and a point $z_0$, suppose that
(a) $f$ is analytic at $z_0$;
(b) $f(z_0) = 0$ but $f(z)$ is not identically equal to zero in any neighborhood of $z_0$.
Then $f(z) \neq 0$ throughout some deleted neighborhood $0 <|z-z_0|<\epsilon$ of $z_0$.
(Brown, J. and Churchill, R., 2014. Complex Variables And Applications. 9th ed. New York: McGraw-Hill Education, p.249.)
This is followed by a proof using the Taylor expansion of $f$ at $z_0$. According to my understanding the authors mean to say that if a function is analytic and not the zero function, then all of its zeros are isolated (an equivalent proof can be found online).
My question is now: Is the formulation of the theorem above not an empty statement, i.e. does $f$ being "not identically zero in any neighbourhood of $z_0$" not mean that $f(z) \neq 0$ everywhere else (where $z \neq z_0$)? Is the conclusion therefore not included in the premise already. Furthermore, is the premise not much too strong, because we might want to allow for other zeros to exist (but isolated from $z_0$).
Sorry, if this question is a bit vague, because the theorem is true and my confusion results entirely from what I suppose it is meant to say. However, there might be some misconception on my side and I hope that someone could point it out.
Best Answer
Asserting that $f$ is identically zero in a neighbourhoood $V$ of $z_0$ means that for all $z\in V$, $f(z)=0$. Therefore, asserting that $f$ is not identically zero in any neighbourhood of $z_0$ means that, for any neighbourhoood $V$ of $z_0$, there is some $z\in V$ such that $f(z)\ne0$.
In order to see that the conclusion is not included in the premise already, consider, say,$$f(z)=\begin{cases}\sin\left(\frac1z\right)&\text{ if }z\ne0\\0&\text{ if }z=0.\end{cases}$$Then $f$ is not identically zero in any neighbourhood of $0$, but $0$ is not an isolated zero of $f$.