Questions: Jordan Curve Theorem (Homological Proof)

Question 1

The Jordan curve theorem seems obvious. Why is a rigorous proof difficult?

• A. Because continuous curves can be arbitrarily pathological — space-filling curves, curves with positive area, wildly oscillating curves — and the theorem must hold for all of them
• B. Because the theorem is actually false for some curves
• C. Because the plane has complicated topology
• D. Because the definition of 'inside' is ambiguous

Answer: (A)

The difficulty lies in the generality of 'continuous simple closed curve.' Such curves can be extremely wild: they can have infinite length, positive Lebesgue measure (the Osgood curve), or oscillate infinitely often at every scale (like space-filling curves, though space-filling curves are not injective so they are not simple). The theorem must handle ALL continuous injections of S^1 into R^2, including those that defy geometric intuition. For smooth or polygonal curves, the theorem is much easier. The homological proof handles all cases uniformly because homology is defined for arbitrary continuous maps.

Question 2

The generalization to higher dimensions (Jordan-Brouwer separation theorem) states: any embedding of S^{n-1} in S^n separates S^n into exactly two components.

• T. True
• F. False

Answer: True

This is the natural higher-dimensional generalization. An embedded (n-1)-sphere in S^n divides S^n into two open connected components whose common boundary is the embedded sphere. The homological proof generalizes cleanly: using the Mayer-Vietoris sequence on the decomposition S^n = U ∪ V where U is the complement of a closed disk neighborhood and V is a neighborhood of the embedded sphere, one computes H_0(S^n \ S^{n-1}) and shows it has rank 2 (two connected components). The proof uses Alexander duality in its most general form.

Question 3

The homological proof of the Jordan curve theorem uses Alexander duality. What does Alexander duality say in this context?

• A. H_k(S^n \ K) ≅ H̃^{n-k-1}(K) for any compact subspace K ⊂ S^n
• B. The fundamental group of the complement equals the homology of the curve
• C. The curve and its complement have the same Euler characteristic
• D. Every closed curve in S^2 is homologous to zero

Answer: (A)

Alexander duality states H_k(S^n \ K) ≅ H̃^{n-k-1}(K) (reduced Cech cohomology). For the Jordan curve theorem: K = C ≅ S^1 embedded in S^2. Then H_0(S^2 \ C) ≅ H̃^0(S^1) ≅ Z. Since H_0 counts connected components minus one (in reduced form), H̃_0(S^2 \ C) ≅ Z means there are exactly 2 components. Alexander duality reduces the 'separation' question (about the complement) to a cohomology computation of the embedded object itself.

Question 4

A figure-eight (two circles touching at a point) in the plane divides the plane into three regions. Explain why this does not contradict the Jordan curve theorem.

Model answer: The Jordan curve theorem applies only to SIMPLE closed curves — continuous injections of S^1 into R^2. A figure-eight is not a simple closed curve: it is the image of a curve that passes through the intersection point twice, so the parametrizing map S^1 → R^2 is not injective at that point. The figure-eight is instead the image of a continuous map that is not an embedding. The theorem makes no claim about non-simple curves, and indeed a non-simple curve can divide the plane into any number of regions.

This is a common source of confusion. The power of the Jordan curve theorem lies in its universality for simple curves — it applies even to incredibly pathological continuous injections. But the simplicity (injectivity) condition is essential. Without it, curves can self-intersect and create arbitrarily many regions.