## A Proof of Liouville’s Theorem in complex analysisAugust 31, 2009

Posted by Phi. Isett in Uncategorized.
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Pretend you had a non-constant, bounded holomorphic function $f$ on all of $\mathbb{C}$.  Since $f$ is bounded at $\infty$, Riemann’s theorem on removable singularities implies that $f$ extends to a holomorphic (and hence continuous) function on the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, which is compact.  If $f$ were not constant, the open mapping theorem would apply to this extension, and the image of the Riemann sphere would be an open subset of $\mathbb{C}$.  But this cannot be the case, because $\mathbb{C}$ has no nonempty subsets which are both open and compact.  $\Box$