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I was going through this website. I am not understanding the definition of a simply connected domain, it says "A simply connected domain is a path-connected domain where one can continuously shrink any simple closed curve into a point while remaining in the domain "

I thought I understood it and my understanding went well with the bellow 2D domains:

2d domain

I can understand that a closed loop path can be shrunk to a point and still be in the domain for the left figure but not for the right one because if it's shrunk to a point then it will breach the inner boundary and form a point inside the inner boundary, which is not in the domain. (Please correct me if my understanding is wrong)

But now when I see the bellow 3D domains, I get confused.

enter image description here

I don't understand why the $2^{nd}$ figure (A sphere having a hollow spherical region) from the left is simply connected. There is a small hollow sphere ( out of domain region) at the centre so if I try to shrink a closed curve (not just any curve but a big circle with radius 99% of the radius of the sphere which is enclosed in the sphere) won't it shrink to a point that's inside the hollow sphere (which is out of the domain)?

Note: The 3D figures with the caption "Non-simply connected" have holes that are drilled all the way through.

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paulplusx
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1 Answers1

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I think your problem is the definition of "shrink to a point", which is more correctly called "homotopic to a constant curve". The "shrinking" curve does not have to lie within the convex hull of the original curve, but is allowed to move around. You can take your big circle and just shrink it while moving it upwards so it never hits the small sphere while you shrink it to a point.

Kusma
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