TITLE: Can a non-constant analytic function have infinitely many zeros on a closed disk?
QUESTION [6 upvotes]: I think not, however my proof is quite sketchy so far..
My attempt: Suppose an analytic function f has infinitely many zeros on some closed disk D. Then there exists a sequence of zeros in D with a limit point in D. Thus by the identity theorem (Let D be a domain and f analytic in D. If the set of zeros Z(f) has a limit point in D, then f ≡ 0 in D.), f is identically zero and thus constant.
My main reasons for confusion (other than having a weak understanding of the identity theorem):
-Couldn't such a function f have a finite number of distinct zeros, each with infinite multiplicity? in this case there wouldn't be a convergent sequence of zeros...
-What is the relevance of the fact that D is closed?
Any help in understanding this problem would be greatly appreciated! 
Thanks

REPLY [3 votes]: Since the disk $\mathcal{D}$ is bounded and closed it is compact, so, by Heine-Borel it fulfils the Bolzano-Weierstrass property: every infinite set has at least one limit point. So the set of $\mathcal{Z}$ of zeros has a limit point, and you're right to go with the identity theorem. So you can see the reason for closure: it lets you call on the Heine-Borel theorem.
Incidentally, before about 1930 and Pavel Alexandrof's school, "compact" meant by definition having the Bolzano-Weierstrass property.