Kodi Arfer / Writing / Introduction to Chaos

Continuous Chaotic Functions Are Sensitive

If (X,d) is an infinite metric space and f:XX is continuous and chaotic, then f sensitively depends on initial conditions.

This proof is from Adams and Franzosa (2008), who in turn got it from Banks, Brooks, Cairns, Davis, and Stacey (1992).

Idea: We find a pair of disjoint periodic orbits O(q) and O(q), so any xX will be at least a constant distance away from at least one of them. We use the first property of chaos (density of periodic points) to get a periodic point p nearby x (call its period m), and the second property of chaos (topological transitivity) to get a point w nearby x such that f eventually brings w near q or q. The continuity of f lets us pick w such that m iterations of f on w stay close to O(q) or O(q). Then for the n we need to demonstrate sensitive dependence on initial conditions, we pick an integer that's a multiple of m but also keeps fn(w) near O(q) or O(q). Then d(fn(p),fn(w))=d(p,fn(w)), which is large because p is near x and x is far from O(q) or O(q). By the triangle inequality, at least one of d(fn(x),fn(p)) or d(fn(x),fn(w)) must be large, proving the theorem.

Proof: First of all, observe that X contains infinitely many periodic points of f, since it's infinite and periodic points are dense in it. At least two of these points—call them q and q—must have disjoint orbits. Otherwise, all of the infinitely many periodic points would be part of the same cycle, prohibiting any of them from having a finite period. Now let δ0=min{d(x,y):xO(q),yO(q)}, so that δ0 is the minimum distance between the orbits of q and q.

[Diagram of the proof so far]

Define δ=δ08. We'll prove that this is the δ needed for sensitive dependence on initial conditions. So pick an arbitrary xX and ε>0. Without loss of generality, we may force ε<δ. (We can shrink ε as much as we like because if we find a suitable yBε(x) for the new ε, then of course yBε(x) for the original, larger ε as well.) Since periodic points are dense, there's a periodic point pBε(x). Let m be the period of p.

It follows from the triangle inequality (and it's obvious from the picture) that wherever p is, it must be at least δ02 units away from either every point in O(q) or every point in O(q); suppose, without loss of generality, that the former is the case.

[Diagram of the proof so far]

For each j=0,1,,m, let Bj=Bδ(fj(q)). (This means that B0 is just Bδ(q). But Bm need not equal B0, since m is the period of p, not q.) Then for each j, the continuity of f implies that fj is continuous, so (fj)-1(Bj) is open; in fact, it's a neighborhood of q. Let V be the intersection of all m+1 of these neighborhoods.

[Diagram of the proof so far]

Since f is topologically transitive, there exist wBε(x) and k such that fk(w)V.

[Diagram of the proof so far]

Now pick h with khmk+m. Suppose we knew that


The triangle inequality would then imply that d(fhm(x),fhm(p))>δ or d(fhm(x),fhm(w))>δ, proving the theorem, since p,wBε(x). So all we have left to do is prove (1).

By the triangle inequality,


Since pBε(x),


Since fk(w)V,


in particular, since khmk+m, fhm(w)Bhm-k=Bδ(fhm-k(q)), so


Combining (2), (3), and (4) yields


But we assumed that x is more than δ02=4δ units away from every point in O(q), so


and thus


This implies (1), since


m being the period of p. □