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Draw two balls centered at the origin in R^3, where $r_1 < r_2$. What is the topological name for the boundary created? Does it even make sense to talk about a figure with an inside boundary and an outside one?

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It's called a "spherical shell", the generalization of the annulus to 3 dimensions. – Arturo Magidin Dec 16 '10 at 20:59
@ Arturo: Thats what I thought, but isn't the spherical shell the set of points in between the boundaries? I'm looking for the set of points that is defined by the surface, much the same way a sphere is defined as not the points inside a certain radius, but the set of points equal to that radius. – Hooked Dec 16 '10 at 21:07
It is a union of two spheres. The fact that they are nested is not an intrinsic property of the boundary itself, just of the particular embedding of the boundary in $\mathbb R^3$. – Grumpy Parsnip Dec 16 '10 at 21:11
You said "balls", not "spheres", so I assumed you meant the filled-in object. If they are spheres, then as Jim Conant says, it's just two disjoint spheres. – Arturo Magidin Dec 16 '10 at 21:15
Yes, in a certain sense it does make sense to talk about a figure with an inside boundary and an outside one. See – Rasmus Dec 16 '10 at 22:41
up vote 3 down vote accepted

Please let me know if the following answer is at the correct level. Thanks!

If you have a "nice" subset $X \subset R^3$ then it makes sense to talk about the outer boundary of $X$ and the inner boundaries. Here "nice" means a connected, compact submanifold of dimension three.

Let's break this down into easy pieces.

If $F \subset R^3$ is a compact, connected surface without boundary then $F$ cuts $R^3$ into exactly two pieces $A$ and $B$ so that the closure of $A$ is compact and the closure of $B$ is not compact. Let's refer to $A$ as the piece "inside" of $F$ and to $B$ as the piece "outside" of $F$. (For help with the exercise please see Hatcher's notes on three-manifolds.)

Useful Lemma: If $F$ and $F'$ are surfaces (as above), and $F$ meets the outside of $F'$, then $F'$ meets the inside of $F$. Proof: Exercise.

Now suppose that $X$ is as above. Let $F_0, F_1, \ldots, F_k$ be the connected components of $\partial X$, the boundary of $X$. Let $A_i$ and $B_i$ be the inner and outer pieces of $R^3 - F_i$. Say that $F_i$ is "inside" $X$ if $X$ meets $B_i$ (and so is disjoint from $A_i$). Otherwise say that $F_i$ is "outside" of $X$. (Exercise: check everything is well-defined using connectedness of $X$!)

Theorem: exactly one of the $F_i$ is outside $X$ and the rest are inside. Proof: Exercise.

Ok, this is a long row to hoe. But this gives precise meanings to inside and outside and then verifies your intuition about those words. It is also comforting that it all comes down to understanding just a few facts -- the Schoneflies theorem (Alexander's theorem in dimension three) and properties of connected sets.

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What an excellent answer! Thank you for the notes, I'll be reading them over. – Hooked Dec 17 '10 at 15:33

Does a plane have two sides? Can the hole be enlarged and stretched so that the inner side of the sphere is everted? The resultant structure should be equivalent to a disc, with Euler characteristic of one. However, the surface is probably polarized, with opposite curl values on either side. this would reflect the opposite vector fields on the spherical surfaces, when viewed from the perspective of an outside or inside observer, relative to the sphere. so without vector fields you don't know which end is up. This is a state characteristic of human beings at their present stage of development.

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