Discussion
Let S be the set of zeros of a non-zero entire function. Then the set S is :
Answer: C
The zero set S is closed.
S is the inverse image of the closed set {0} under the continuous function \(f\), so it is closed. More is true: because \(f\) is not identically zero, its zeros are isolated, so S is a discrete closed set with no limit point in C.
The other descriptions fail. S cannot contain an open set, since then \(f\) would vanish identically. It cannot be infinite and bounded, since a bounded infinite set has a limit point, which would force \(f \equiv 0\). And a discrete subset of C is countable. S may even be empty, as for \(e^{z}\).
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