Corollary of identity theorem without connectedness assumption The identity theorem has a corollary, which is often stated as "If $U$ is a connected domain, $f,g$ are analytic in $U$, and the set of points where $f$ and $g$ coincide has a limit point in $U$, then $f=g.$"
The proof runs by showing that the set $L$ of limit points of $\{z \in U : f(z) = 0\}$ is closed and open. Since $U$ is connected,  $L$ is either empty or all of $U$. Since $L$ is nonempty (there is a limit point in $U$), we conclude that $L = U$; that is, $f(z) = 0$ on all of $U$. By replacing "$f(z)$" with "$(f-g)(z)$", the conclusion follows.
However, I have also seen this corollary stated withput the connectedness assumption; that is, "If $f,g$ are analytic on some $U$, and the set of points where $f$ and $g$ coincide has a limit point in $U$, then $f=g.$" Wikipedia seems to confirm this in the "An Improvement" section: http://en.wikipedia.org/wiki/Identity_theorem
However, I'm having a hard time proving the corollary without the usual assumption of connectedness. Also, I don't understand Wikipedia's proof, because it just seems to show (in the end) that $f$ and $g$ must be equal in a neighborhood of the limit point of $U$ (by showing that $f^k(c) = g^k(c)$ for all $k \ge 0$, where $c$ is the limit point).
Any help would be appreciated. 

Where have you seen it stated without connectedness?. For what i understand, A domain = an open connected set. "Specifically, if two holomorphic functions f and g on a domain D agree on a set S which has an accumulation point c in D then f = g on all of D.". You're still using connectedness but instead of f and g being equal on a whole subset they're equal on a sequence with a limit point.
http://en.wikipedia.org/wiki/Domain_%28mathematical_analysis%29
