# Prove that the set of those $c$ is countable

by zeraoulia
Tags: analytic function, real and complex
 P: 12 The **mean values theorem** says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u)$$ My question is: Assume that $u$ is a root of $f$, hence we obtain $$f(v)=f′(c)(v-u)$$ Assume that $f$ is a non-zero analytic function in the whole real line. My interest is about the real $c∈(u,v)$. I believe that the set of those $c$ is contable, otherwise we conclud that $f′(c)$ is constant (in an open set containing $c$) since $v$ is a constant and hence $f$ is identically zero. Also, I think that the set of those $c$ is countable (for analytic functions) but I am not able to prove that.
 P: 178 For all c that satisfy that equation, $f'(c)(v-u)-f(v)=0$. The LHS is an analytic function in c (v and u are constant) and is not identically zero unless f is linear. That any non-zero analytic function has countable zeroes is well known. (http://planetmath.org/ZeroesOfAnalyt...onsAreIsolated is a proof for complex functions, but the idea is the same) Edit: My link only proves that the zeroes are isolated. Here is how you would show that the set of zeroes in countable: Let a be a zero. If all other zeroes are separated from a by at least ε, let q be a rational number which is no more than ε/3 away from a. Each rational number can be associated with a zero in this way. The cardinality of the set of zeroes is the same as for this set of rational numbers and rational numbers are countable.
 PF Patron Sci Advisor Emeritus P: 8,837 Type ## instead of \$.

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