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"Every bounded function that is holomorphic on $A$ is is constant."

For which $A\subseteq\mathbb{C}$ is this true?

Are there well-known examples of unbounded sets $A\subseteq\mathbb{C}$ on which there are non-constant bounded holomorphic functions?

Later edit: My striking through the second question was meant only to de-emphasize it. Feel free to post further on it if you wish. Some of the examples posted in response to it were already well known to me; I'd have thought of them if my attention had been on the second question rather than the first.

I'm envisioning a couple of possibilities: (1) Various other sorts of sets $A$ will be mentioned in answers; and (2) An answer will say that some nice theorem says this is true of a set $A$ if and only if whatever, where "whatever" is something non-trivially different from a tautologous "if and only if every bounded holomorphic function on $A$ is constant", and maybe "whatever" is somehow elegant or at least simple.

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    You might want to assume something like connectedness because then you can have a "constant by parts" function on the different connected components. But this is a rather trivial remark.2012-08-11
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    What do you mean by holomorphic if $A$ isn't open?2012-08-11
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    @QiaochuYuan : Possibly an interesting question. Maybe I should just ask for which _open_ sets $A\subseteq\mathbb{C}$ this is true. But I see someone's posted an answer involving a set that is not open. I'm not sure what to make of that example yet.2012-08-11
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    I see plenty of answers to my second question. I'm not surprised; it seems like the easy part. The answer from Jose27 seems to attempt to address the first question, and I'm not sure I understand it yet.2012-08-11
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    $@$Michael: the only reasonable definition of a holomorphic function on an arbitrary subset $A$ of $\mathbb{C}$ that I've seen is: $f$ extends to a holomorphic function on some open subset $U \supset A$. Thus you may as well assume that $A$ is open.2012-08-11
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    By the way, this is a very interesting question. It deserves more than the +4 it currently has.2012-08-12
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    The answer is: precisely for the sets such that $\mathbb C\setminus A$ is removable for bounded holomorphic functions. Which leads to [another question](http://math.stackexchange.com/questions/159659/which-sets-are-removable-for-holomorphic-functions).2012-08-13

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