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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Correction to 'meromorphic'
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closedness - still by cgs
Correction id: 14661
Filed on: 2009-06-03 20:13:09
Status: Rejected on 2009-11-23 03:20:20
Type: Erratum
Correction text:
>Your suggested correction leads to an equivalent definition, because one >can prove that a function with an isolated set of poles and no other >singularities has the property that the set of its poles is closed. >Indeed, this theorem is even mentioned in the text of the entry itself.
In the definition the set of poles has to be closed, for otherwise the notion of f being homomorphic on the complement would not be defined |
Comment from object owner djao:
> In the definition the set of poles has to be closed, for otherwise the
> notion of f being homomorphic on the complement would not be defined
Since the definition of meromorphic specifically requires that f is holomorphic on the complement of the poles, it also (a fortiori) requires that the notion of f being holomorphic on the complement of the poles is defined. Therefore it is not necessary to include in the definition that the set of poles must be closed. The closedness of the set of poles is implied by the requirement that f is holomorphic on the complement of the poles.
To put it another way, let f: U \to C be a function with an isolated non-closed set of poles P \subset U, and no other poles. Let C be the closure of P. We may certainly take f to be holomorphic at every point in U \setminus C (this is well defined, since U \setminus C is open).
Claim: For any point z \in C, there exists no neighborhood of z on which f is holomorphic. Indeed, there exists no neighborhood D of z for which f is defined at every point in D, since D \cap P is nonempty.
Claim: f does not have a pole at any point on C \setminus P. Indeed, let z \in C \setminus P. Then there exists no deleted neighborhood D of z for which f is defined at every point in D \setminus {z}, since (D \setminus {z}) \cap P is nonempty.
Claim: f is not meromorphic. Let z \in C \setminus P. Such a point z exists, since P is non-closed. Then f is not holomorphic at z (or on any neighborhood of z), but f does not have a pole at z. However, the definition of meromorphic requires f to be holomorphic on (a neighborhood of) every point at which it does not have a pole. Since the point z fails this requirement, f is not meromorphic. |
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