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Great Picard theorem
Statement
Let be holomorphic on a punctured disc and suppose is an essential singularity of . With at most one finite exception, every value in is assumed infinitely often in every punctured neighborhood of .
Facts & Assumptions
Given: A holomorphic function on with an essential singularity at .
If a punctured-disc holomorphic function omits two distinct finite values, then the singularity is removable or a pole (Two omitted finite values rule out an essential singularity).
Proof
Suppose two distinct finite values each failed to occur infinitely often in some punctured neighborhood of . After passing to the smaller of those neighborhoods, each equation would have only finitely many solutions there. Shrink once more past all those finitely many points. The resulting punctured disc omits both and , so [L1] would make the singularity removable or a pole, contradicting the hypothesis that it is essential.
Step 1.1 shows that at most one finite value can fail the asserted infinitely-often property. Every other finite value is therefore assumed infinitely often in every punctured neighborhood of .
This is exactly the Great Picard conclusion for finite values.
Depends on
Used by
Dependency tree · two levels
8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Aleksander Simonic, The Ahlfors lemma and Picard's theorems, §6.4 (standard reference, not scraped)