- #1
Jamin2112
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- 12
Homework Statement
As in in title.
Homework Equations
Open mapping: maps open sets to open sets.
The Attempt at a Solution
Not sure.
Jamin2112 said:Homework Statement
As in in title.
Homework Equations
Open mapping: maps open sets to open sets.
The Attempt at a Solution
Not sure.
Dick said:That's a pretty poor attempt. If you don't know how to prove it can you at least give us a few thoughts on why you think it might (or might not) be true?
Jamin2112 said:If a function f(z) is differentiable on a set G, then it is continuous on G. If it continuous, the image of every open set is an open set. Done(?).
Dick said:Nice that you are trying to think about it. But that's not true in the real numbers. f(x)=x^2 is continuous and differentiable, but it's not open. The proof is going to have to involve special properties of analytic functions.
Jamin2112 said:Anything to do with "Analytic continuation" that I see on Wikipedia?
An analytic mapping is a function between two complex manifolds that is infinitely differentiable. This means that the function has derivatives of all orders at every point in its domain.
An open mapping is a function that preserves the open sets of a topological space. In other words, if an open set is mapped to another space, the image of the open set will also be open.
To show that an analytic mapping is an open mapping, you must prove that the function preserves open sets. This can be done by showing that the image of any open set in the domain is also open in the codomain.
An analytic mapping being an open mapping means that the function is well-behaved and has nice properties. It also allows for the use of complex analysis techniques to study the function.
Yes, it is possible for a mapping to be analytic but not open. This can occur if the function fails to preserve open sets in its domain or if the function is not defined on a complex manifold.