I have an idea how to do this, but I am a noob in topology.
Given a contour . Can a circle be "shrunk" to be strictly contained in the interior ?
Let be a complex function defined on an open non-empty set . Let be a contour wholly inside . Does there exist another contour wholly inside with the property (abusing notation) that (meaning such that this contour lie strictly within this new contour).
Yes, let be so that:
1) is a path, i.e. the components of (IM and RE) are continous.
2) i.e. it is closed.
3) , i.e. it is simple.
Is that supposed to be a pun! No, seriously I think you and I have a different understanding of what a contour is. See above.Originally Posted by topsquark
The Jordan Curve Theorem states that any simple closed curve partitions the plane into three non-empty disjoint connected sets: the curve itself, a bounded set (the interior) and an unbounded set (the exterior). An arc from an interior point to an exterior point must contain a point of the curve.
Moreover, to answer your particular question: yes about any interior point there is a closed disk (can be centered at the point) that is a subset of the interior.
Actually I did not answer anything general about contours. I only answer what TPH was considering a contour, for him it is a simple closed curve. That is by no means standard. Most complex variables textbooks would consider a contour as the image of a continuous one-to-one mapping of [0,1]: a Jordan arc. If we do allow only f(0)=f(1) the we have a simple closed curve.
The original set must be open. If that is the case, then you can inscribe a circle inside all contours C contained in the set.
To contruct it, well...
- Jordan's theorem gives us that the interior of C is open, so there certainly are circles everywhere. Here, we might also drop connectedness (though topsq made an important point).
On the other hand, that is some big theorem, and you don't normally call a bulldozer to strike a nail. There is a procedure that can produce a new curve inside this one, and more importantly - maintain the same winding number on the complement of S (as, in the important cases, S has holes where functions have poles). Take the curve to be C(t), t in [0,1]. Let n(t) be the normal to c(t), pointing inwards (since C is simple, this is always well defined). Now there is a s>0, such that sn(t) is wholly inside S (as this is open) and the contour
C*(t)=C(t)+sn(t), t in [0,1]
is a) inscribed in C and b) for a function f, I(C,f)=I(C*,f) on the compliment of S, so the Cauchy theorems apply.
I think I finally understand, thank you!
I was discussing this with my professor by e-mail. He said that the book by Lang makes the following definition:
Definition: Given a contour a point is "inside" iff . (I really like it).
So we define to be all points in such that the contour integral over Western Europe is non-zero.
Now I think I understand, "open" means we can find a such that (the neighorbhood) . So the circle is what we seek. And this is possible by the Jordan Closed Curve theorem which Plato explains as saying that is an open set.