Let $f:ℝ→ℝ$ be a real analytic function. If $f$ has infinitely many zeros, then we know that the fiber $f^{-1}(0)$ is an infinite discrete and countable set. Let $a∈ℝ,a≠0$, we know also that the fiber $f^{-1}(a)$ is a discrete set unless $f = a$. Then my question is: Some observations show that the number of the fibers $f^{-1}(a)$ is finite (not the number of elements in each fiber). However, I am not able to prove that. We can define the equivalence relation as: $$xRy⇔f(x)=f(y)$$
The equivalence classes of the relation $R$ are the fibers $f^{-1}(x)$. If the number of those fibers is infinite, then this will create a contradiction with the fact that $f$ is an analytic function.
Summarizing the discussion in comments: the fibers of $f$ are in bijective correspondence with the points in the range of $f$. The range of a nonconstant continuous real function is uncountable.
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