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It is a basic fact that the zero set of a non zero holomorphic function defined on a open set $A$ is discrete.

By a result in Rudin's textbook on "Real and Complex Analysis", we know that any real valued harmonic function $u$ is locally the real part of some holomorphic function, so I ask the following question:

Is the zero set $Z(u)$ of a non zero real valued harmonic function $u$ defined on open set $A\subset \mathbb{C}$ discrete?

Note that if $u$ vanishes on a nonempty open set $O\subset A$, then write $f=u+iv$ to be the holomorphic function, then Cauchy-Riemman's theorem implies that $v=0$ on $O$, so $f=0$ on $O$, then $f=0$ on $A$.

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Virtually every harmonic function is a counterexample... – mrf Jan 8 '13 at 22:52
up vote 2 down vote accepted

No. The function $f(a+ib)=b$ is harmonic but its zero set is the real line.

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