Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 16 / 2

The convention intended by this question is existence of a dense forward orbit, also called point transitivity. On a nonempty compact metric space, this means that some $x\in X$ satisfiesEquivalently, that forward orbit visits every nonempty open set.

The other common definition of topological transitivity uses intersections of iterated open sets, as in part (b). The distinction matters on spaces with isolated points: parts (b) and (c) ask precisely how the two conventions are related. We use point transitivity for the dense-orbit convention throughout this solution. As usual for this definition, the state space is assumed nonempty; on an empty space the open-set condition is vacuous and there is no point with a forward orbit.

A compact metric space is second countable. To see this concretely, for each positive integer $r$ choose a finite cover by balls of radius $1/r$. Their centres, over all $r$, form a countable dense subset; balls centred there with positive rational radii form a countable basis of a topology. Enumerate its nonempty members as $(V_j{)}_{j\ge 1}$, repeating members if the basis is finite.

For each $j$, putEach $G_j$ is an open set, because all iterations of a map $f^n$ are continuous. It is dense: if $U$ is any nonempty open set, the hypothesis supplies $n\ge 1$ and $u\in U$ with $f^n(u)\in V_j$, so $u\in U\cap G_j$.

A compact metric space is a complete metric space, hence a Baire space. The Baire category theorem implies that $G={\bigcap }_{j\ge 1}{G}_j$ is dense, and in particular nonempty. For any $x\in G$, the forward orbit of $x$ meets every $V_j$, and therefore every nonempty open set. Its orbit closure is $X$. Thus the open-set intersection hypothesis implies point transitivity, and in fact the set of points with a dense forward orbit contains the dense intersection $G$.

Let $x$ have a dense forward orbit $O=\{f^j(x):j\ge 0\}$, and suppose that $f$ is surjective. Every tail $O_k=\{f^j(x):j\ge k\}$ is then dense. Indeed, continuity givesand the left side equals $f^k(X)=X$ by surjectivity.

Given nonempty open sets $U,V$, choose $k\ge 0$ with $f^k(x)\in U$. The dense tail $O_{k+1}$ supplies $j\ge k+1$ with $f^j(x)\in V$. Taking $n=j-k\ge 1$ gives $f^j(x)\in f^n(U)\cap V$. This proves the required positive-time intersection property.

The result can fail without surjectivity. Take the compact metric spacewith the metric induced from the real line. All nonzero points are isolated, so continuity there is automatic; at zero, $f(1/j)\to 0=f(0)$ proves continuity. The forward orbit of $1$ is $\{1,1/2,1/3,\dots \}$, whose closure is $X$, so this is point transitivity. But $f(X)=X\setminus \{1\}$. The singletons $U=\{1/2\}$ and $V=\{1\}$ are nonempty open sets, and for every $n\ge 1$,The isolated initial point can be visited only at the start of the dense forward orbit; deleting that initial segment destroys density.