What Makes the Jordan Curve Theorem a Cornerstone of Topology?

[Silviu]

Hello! I came across Jordan Curve Theorem while reading something on Complex Analysis. I don't know much about topology and I apologize if my questions is silly, but from what I understand the theorem states that a closed curve in the complex plane separate the plane into an inner region and an outer region whose boundary is the curve. I read that the theorem has a rich history and it was even checked using computer programs, but to me it seems quite obvious. What exactly am I missing about it (why it is considered so important)? Thank you!

 

[fresh_42]

Yes, it's obvious and it has been used for years without any proof. But only because something appears to be obvious, that doesn't mean it's also true or proven. E.g. it's obvious that you cannot paint a square with a line, since it has only one dimension. False, you can. So I suggest to try and prove it yourself and see whether it remains "obvious".

 

[WWGD]

Hello! I came across Jordan Curve Theorem while reading something on Complex Analysis. I don't know much about topology and I apologize if my questions is silly, but from what I understand the theorem states that a closed curve in the complex plane separate the plane into an inner region and an outer region whose boundary is the curve. I read that the theorem has a rich history and it was even checked using computer programs, but to me it seems quite obvious. What exactly am I missing about it (why it is considered so important)? Thank you!

I think you may be considering only curves that are very straightforward and not those curves that can be very complicated, where it is difficult to tell apart the interior of the curve from its exterior.

 

[WWGD]

Yes, it's obvious and it has been used for years without any proof. But only because something appears to be obvious, that doesn't mean it's also true or proven. E.g. it's obvious that you cannot paint a square with a line, since it has only one dimension. False, you can. So I suggest to try and prove it yourself and see whether it remains "obvious".

Space-filling curves?

 

[Silviu]

I think you may be considering only curves that are very straightforward and not those curves that can be very complicated, where it is difficult to tell apart the interior of the curve from its exterior.

Thank you for your reply. Could you please point me towards such a curve?

 

[lavinia]

Thank you for your reply. Could you please point me towards such a curve?

There are curves that do not have a tangent line at any point. A classic one of these that is homeomorphic to a circle is the Koch snowflake.

In the plane more is true. A continuous non-self intersecting closed curve not only divides that plane into two regions of which it is the mutual boundary but there is a homeomorphism of the plane that maps the curve to a circle and the two regions to the interior of the circle and the unbounded outside of the circle. This is not true in higher dimensions for instance for homeomorphs of the 2 sphere embedded in 3 space.

 

[WWGD]

There are curves that do not have a tangent line at any point. A classic one of these that is homeomorphic to a circle is the Koch snowflake.

In the plane more is true. A continuous non-self intersecting closed curve not only divides that plane into two regions of which it is the mutual boundary but there is a homeomorphism of the plane that maps the curve to a circle and the two regions to the interior of the circle and the unbounded outside of the circle. This is not true in higher dimensions for instance for homeomorphs of the 2 sphere embedded in 3 space.

Yes, I think this is equivalent to the existence of knotted spheres, isn't it? Do you remember the name of this theorem? I have it on the tip of my tongue ( the name, not the theorem ;) ).

 

[lavinia]

Yes, I think this is equivalent to the existence of knotted spheres, isn't it? Do you remember the name of this theorem? I have it on the tip of my tongue ( the name, not the theorem ;) ).

I don't know anything about knotted spheres. The first known example was Alexander's horned sphere. I don't think that the non-compact component of its complement is even simply connected.

A powerful theorem that one can use here is Alexander Duality. Alexander Duality works for any compact manifold in the n-sphere. For a closed hyper-surface the theorem says that the sphere is split into two components.

 

[WWGD]

I don't know anything about knotted spheres. The first know example was Alexander's horned sphere. I don't think that the non-compact component of its complement is even simply connected.

Yes, this , the non-simple-connectedness of the sphere , is the obstruction to the outside of the boundary surface mapping to the 2-sphere; the complement of the standard 2-sphere is simply-connected, but that of the Alexander horned sphere is not.

 

[lavinia]

Yes, this , the non-simple-connectedness of the sphere , is the obstruction to the outside of the boundary surface mapping to the 2-sphere; the complement of the standard 2-sphere is simply-connected, but that of the Alexander horned sphere is not.

Not sure what you mean, The sphere is simply connected except for the circle.

 

[WWGD]

Not sure what you mean, The sphere is simply connected except for the circle.

I mean the complement of the standard sphere is simply-connected, but the complement of its homeomorph the Alexander horned sphere, is not, which creates a counter in 2-D , i.e., for 2D-spheres, of the theorem you cited.

EDIT: The theorem I was referring to is the Jordan-Schoenflies theorem https://en.wikipedia.org/wiki/Schoenflies_problem, which is the one you also referred to.: every simple-closed theorem separates the plane into two regions, each respectively homeomorphic to the interior and the exterior of the circle. The Alexander Horned sphere is a counter to a generalization : a "simple" ( non-intersecting, I guess ) surface does not divide the plane into two regions with similar properties; specifically, the exterior of the horned sphere is not homeomorphic to the exterior of the standard sphere, since one of them is simply-connected, while the other one is not (I think homotopy - equivalence itself is enough to guarantee simple-connectedness).

 

[lavinia]

Ok.

I mean the complement of the standard sphere is simply-connected, but the complement of its homeomorph the Alexander horned sphere, is not, which creates a counter in 2-D , i.e., for 2D-spheres, of the theorem you cited.

OK. Just checking.

This example shows that Euclidean space is more complicated that it first appears. One can think of it as a simple contractible n-dimensional extension but that misses a world of structure.

BTW: I think Alexander Duality with $Z_2$ coefficients can be used to show that a smooth closed hyper surface of Euclidean space must be orientable.

 

[WWGD]

Ok.

OK. Just checking.

This example shows that Euclidean space is more complicated that it first appears. One can think of it as a simple contractible n-dimensional extension but that misses a world of structure.

BTW: I think Alexander Duality with $Z_2$ coefficients can be used to show that a smooth closed hyper surface of Euclidean space must be orientable.

Can you think of the general condition for 2D-or-higher of a closed curve? Clearly not just any manifold will do.

 

[lavinia]

Can you think of the general condition for 2D-or-higher of a closed curve? Clearly not just any manifold will do.

Not sure what you mean for a closed curve in dimensions above 2. What is the condition you are looking to satisfy?

 

[lavinia]

The Jordan curve Theorem is always true in any dimension. But not all embeddings of (n-1) spheres in n-space are equivalent. By equivalence is meant: There is a homeomorphism of $R^{n}$ into itself that carries one sphere into the other. All embeddings are equivalent in $R^2$ but in higher dimensions embeddings can be "wild" and not equivalent to the standard embedding.

There is much research on this. Here is a review paper on the topological ideas used in studying this phenomenon for 2 sphere in $R^3$

https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/RHBing.pdf

 

[WWGD]

Not sure what you mean for a closed curve in dimensions above 2. What is the condition you are looking to satisfy?

It would not be a curve, but a surface that I want to have it divide 3-space into two regions, similar to the case of Jordan curve, and , even better, satisfy Schoenflies condition that the interior of the surface maps homeomorphically into the interior of a standard sphere, same for the exterior. So we need a surface with a non-empty (topological) interior.

 

[WWGD]

The Jordan curve Theorem is always true in any dimension. But not all embeddings of (n-1) spheres in n-space are equivalent. By equivalence is meant: There is a homeomorphism of $R^{n}$ into itself that carries one sphere into the other. All embeddings are equivalent in $R^2$ but in higher dimensions embeddings can be "wild" and not equivalent to the standard embedding.

There is much research on this. Here is a review paper on the topological ideas used in studying this phenomenon for 2 sphere in $R^3$

https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/RHBing.pdf

I think this is equivalent to knotedness, when there are non-homotopically ( or other "category") embeddings . Any two embeddings of disks in 2d are equivalent, so there are no knots. Maybe there are other spaces that have more than one inequivalent embedding in the plane.

 

[lavinia]

I think this is equivalent to knotedness, when there are non-homotopically ( or other "category") embeddings . Any two embeddings of disks in 2d are equivalent, so there are no knots. Maybe there are other spaces that have more than one inequivalent embedding in the plane.

I don't think a hyper surface can be a knot.

 

[WWGD]

I don't think a hyper surface can be a knot.

I think, but not sure, that a surface S is knotted in an ambient environment A if there are non-isotopic embeddings of S in A , i.e., if there are embeddings S, S' and no ambient isotopy taking S to S'.

 

[lavinia]

I think, but not sure, that a surface S is knotted in an ambient environment A if there are non-isotopic embeddings of S in A , i.e., if there are embeddings S, S' and no ambient isotopy taking S to S'.

The theory of knots seems to require at least codimension 2 . For instance one can not knot a circle in the plane. Isotopy would seem to require a smooth embedding. But the Alexander horned sphere is not smoothly embedded. (I gather from cursory web surfing that the generalization of a smooth embedding is a "flat" embedding but I am not knowledgeable about this. Certainly the Alexander horned sphere is not flatly embedded. )

I believe that if the sphere $S^{n-1}$ is smoothly embedded as a hyper-surface then of $R^{n}$ then both components of its complement must be simply connected. (The bounded component is always simply connected whether or not the embedding is smooth. So the problem reduces to showing that the unbounded component is simply connected. )

The idea is that a smooth embedding guarantees that there is a tubular neighborhood of the embedded sphere and this neighborhood allows one to apply Van Kampen's Theorem to conclude that the fundamental group of $S^{n}$ is equal to the fundamental group of the unbounded component. In the case of Alexander's horned sphere the unbounded component has an infinite fundamental group.

This suggests that the embedding must be quite weird in order for the unbounded component to have a non-trivial fundamental group.

Why not try to fill in the details of this idea to see if it really works.

 

[Svein]

Thank you for your reply. Could you please point me towards such a curve?

There are several. Check out the Sierpinski curves and the Hilbert curves.

 

[WWGD]

Like Svein says, the circle can be embedded in pretty wild ways beyond the obvious ones one is most likely to run into unless one is doing specialized work/applications. And it gets wilder in higher dimensions, where Jordan curve generalizes, but Schoenflies theorem ( the s.c. curve separates $\mathbb R^2$ into two regions, one homeomorphic to the interior of a circle and the other region homeomorphic to the exterior of the s.c.curve. See Alexander's Horned Sphere , as Lavinia stated, for a counterexample in $\mathbb R^3$ ) does not, i.e., there are knots in higher dimensions.