Continuous Chaotic Functions Are Sensitive

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 $O\left(q\right)$ and $O(q\prime )$, so any $x\in X$ 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 $q\prime$. The continuity of $f$ lets us pick $w$ such that $m$ iterations of $f$ on $w$ stay close to $O\left(q\right)$ or $O(q\prime )$. 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 $f^n\left(w\right)$ near $O\left(q\right)$ or $O(q\prime )$. Then $d(f^n\left(p\right),f^n\left(w\right))=d(p,f^n\left(w\right))$, which is large because $p$ is near $x$ and $x$ is far from $O\left(q\right)$ or $O(q\prime )$. By the triangle inequality, at least one of $d(f^n\left(x\right),f^n\left(p\right))$ or $d(f^n\left(x\right),f^n\left(w\right))$ 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 $q\prime$—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 ${\delta }_0=min\{d(x,y):x\in O\left(q\right),y\in O(q\prime )\}$, so that ${\delta }_0$ is the minimum distance between the orbits of $q$ and $q\prime$.

Define $\delta =\frac{{\delta }_0}{8}$. We'll prove that this is the $\delta$ needed for sensitive dependence on initial conditions. So pick an arbitrary $x\in X$ and $\epsilon >0$. Without loss of generality, we may force $\epsilon <\delta$. (We can shrink $\epsilon$ as much as we like because if we find a suitable $y\in B_{\epsilon }\left(x\right)$ for the new $\epsilon$, then of course $y\in B_{\epsilon }\left(x\right)$ for the original, larger $\epsilon$ as well.) Since periodic points are dense, there's a periodic point $p\in B_{\epsilon }\left(x\right)$. 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 $\frac{{\delta }_0}{2}$ units away from either every point in $O\left(q\right)$ or every point in $O(q\prime )$; suppose, without loss of generality, that the former is the case.

For each $j=0,1,\dots ,m$, let $B_j=B_{\delta }\left(f^j\left(q\right)\right)$. (This means that $B_0$ is just $B_{\delta }\left(q\right)$. But $B_m$ need not equal $B_0$, since $m$ is the period of $p$, not $q$.) Then for each $j$, the continuity of $f$ implies that $f^j$ is continuous, so ${\left(f^j\right)}^{-1}\left(B_j\right)$ is open; in fact, it's a neighborhood of $q$. Let $V$ be the intersection of all $m+1$ of these neighborhoods.

Since $f$ is topologically transitive, there exist $w\in B_{\epsilon }\left(x\right)$ and $k\in \mathbb{N}$ such that $f^k\left(w\right)\in V$.

Now pick $h\in \mathbb{N}$ with $k\le hm\le k+m$. Suppose we knew that

The triangle inequality would then imply that $d(f^{hm}\left(x\right),f^{hm}\left(p\right))>\delta$ or $d(f^{hm}\left(x\right),f^{hm}\left(w\right))>\delta$, proving the theorem, since $p,w\in B_{\epsilon }\left(x\right)$. So all we have left to do is prove (1).

By the triangle inequality,

Since $p\in B_{\epsilon }\left(x\right)$,

Since $f^k\left(w\right)\in V$,

in particular, since $k\le hm\le k+m$, $f^{hm}\left(w\right)\in B_{hm-k}=B_{\delta }\left(f^{hm-k}\left(q\right)\right)$, so

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

But we assumed that $x$ is more than $\frac{{\delta }_0}{2}=4\delta$ units away from every point in $O\left(q\right)$, so

and thus

This implies (1), since

$m$ being the period of $p$. □