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For a continuous map on a compact metric space, the Bowen metric compares the first $n$ points of two forward orbits. Its radius-$ϵ$ ball is $B_n(x,ϵ)=\{y:d_n(x,y)<ϵ\}$. If $s_n(ϵ)$ is the largest cardinality of an $ϵ$-separated set in this metric, then topological entropy is ${\lim }_{ϵ\downarrow 0}{\mathrm{limsup}}_{n\to \infty }{n}^{-1}\log s_n(ϵ)$. The uniform continuity of the identity between two compatible metrics gives the comparison of separated-set counts needed for independence of the original metric.

Given a continuous surjection $h:\Omega \to K$ from Cantor space to a nonempty compact metric space, the pullback $f\mapsto f\circ h$ is a unital isometric embedding of $C(K)$ into $C(\Omega )$. Transport a normalized positive linear functional to its image, take a positive extension from a unital subspace of C(K), and represent that extension on Cantor space. The pushforward measure $h_{\ast }P$ then represents the original functional.

Let $a_m\ge 0$ and $a_m\to 0$. The set $K=\{x=(x_m):x_m\in \mathbb{R},|x_m|\le a_m\}$ is a compact metric space for the supremum norm metric. Every member is a sequence tending to zero.

For any sequence in $K$, successively extract a convergent subsequence in each compact interval $[-a_m,a_m]$, then take diagonal indices. The coordinate limits define $x\in K$. For any $M$, coordinatewise differences in the tail are bounded by $2{\sup }_{m>M}{a}_m$, independently of the sequence index. This tends to zero as $M\to \infty$, while finitely many initial coordinates converge together. Hence the diagonal subsequence converges in the supremum norm. This proves sequential compactness, equivalent to compactness for metric spaces.

On a compact metric space, every sequence of Borel probability measures has a subsequence with weak convergence of probability measures to a Borel probability measure. To see this, choose a countable uniformly dense subset of continuous functions, extract a diagonal subsequence of their bounded integrals, and extend the resulting positive normalized functional to all continuous functions. The Riesz representation theorem supplies its Borel probability measure. This is useful for constructing an invariant measure from orbit empirical measures.

For a Borel probability measure on a compact metric space, inner regularity and outer regularity give $K\subseteq B\subseteq U$ with $K$ compact, $U$ open and $\mu (U\setminus K)<ϵ$. The Urysohn lemma supplies a continuous function $\phi$ equal to one on $K$ and zero outside $U$, with values in $[0,1]$. Its difference from the indicator function of $B$ is supported in $U\setminus K$ and bounded by one, proving the displayed estimate. This transfers identities established for continuous test functions to measurable sets.

The two-sided full shift on a finite alphabet $[k]$ consists of all functions $z:\mathbb{Z}\to [k]$, with the product topology and the left shift. It is a compact metric space. A compatible metric isAgreement on increasingly large finite coordinate sets is equivalent to convergence in this topology. The metric above is compatible but is not invariant under the left shift.

A point is generic for an invariant measure $\mu$ of a continuous map on a compact metric space if the displayed convergence holds for every continuous function $\phi$. Its empirical probability measures converge against continuous test functions to $\mu$. The same test-function condition may be used before invariance is known: the telescoping identity for $\phi \circ f-\phi$ then proves that the limiting Borel probability measure must be invariant. If almost every point is generic for the same measure, continuous approximation of invariant-set indicators proves that this measure is ergodic.

For compact metric spaces $X$ and $Y$, the Gromov-Hausdorff distance iswhere the infimum runs over all metric spaces $Z$ and all isometric embeddings $\phi :X\to Z$ and $\psi :Y\to Z$. It measures how closely the two spaces can be placed inside one ambient metric space.

The Gromov-Hausdorff topology on isometry classes of compact metric spaces is the topology induced by the Gromov-Hausdorff distance.

Two maximum-cardinality $\epsilon$-separated subsets $A,B$ of a compact metric space have a bijection that moves each point by less than $\epsilon$. Join $a$ to $b$ when their distance is below $\epsilon$. If $S\subseteq A$ had fewer than $|S|$ neighbours, replacing these neighbours in $B$ by $S$ would create a larger separated set. The Hall marriage theorem therefore supplies a perfect matching. This allows empirical averages on a compact isometry group to be made approximately invariant without assuming a prior Haar measure construction.

For a measure on a topological space in which compact sets are measurable, inner regularity means that the measure of each measurable set is the supremum of the measures of its contained compact sets. Every Borel probability measure on a compact metric space is inner regular and has outer regularity. Combining the two properties allows approximation of an indicator function by a continuous function in the $L^1$ norm.

Let $X$ be a compact metric space and $\tau :X\to X$ a continuous map. Its composition operator on the Banach space $C(X)$ is $Tf=f\circ \tau$. A continuous linear functional $\mu$ is invariant when $\mu (Tf)=\mu (f)$ for every $f\in C(X)$.

To construct a positive normalized invariant linear functional, fix $x\in X$ and average point evaluations along its orbit:These linear functionals have operator norm one. The Banach-Alaoglu theorem, together with separability of $C(X)$, gives a subsequence convergent in the weak-star topology. Its limit $\mu$ is positive and satisfies $\mu (1)=1$. Finally,so the limit is invariant. Continuity of $\tau$ ensures $Tf\in C(X)$, which is needed when passing to the limit in the weak-star topology.

For continuous functions $f,g$ on a compact metric space satisfying $f(x)+g(y)\le d(x,y)$, set $h(x)={\min }_y(d(x,y)-g(y))$. The triangle inequality makes $h$ $1$-Lipschitz, while the constraint gives $f\le h$ and evaluation at $y=x$ gives $g\le -h$. Thus the feasible pair $(h,-h)$ has at least as large an objective against any two probability measures. This explains why metric optimal transport duality reduces to a single $1$-Lipschitz function.

A cover of $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The equivalent version for open sets follows by shrinking finite open covers on a compact metric space; the reverse implication follows by enlarging antipodal-pair-free closed sets slightly. This is a covering formulation of the Borsuk-Ulam theorem.

Let $X$ be a compact metric space, let $\mu$ be a finite Borel measure whose support of a measure is all of $X$, and let $K$ be a continuous Hermitian positive-semidefinite kernel. The integral operatoris a positive compact operator on $L^2(\mu )$. Its strictly positive eigenvalues ${\lambda }_j$ have orthonormal eigenfunctions with continuous representatives ${\varphi }_j$, and Mercer's theorem givesThe series converges uniformly on $X\times X$ and absolutely at every pair of points. These measure and support assumptions specify where the eigenfunction expansion reconstructs the original kernel.^[ref]

For why full support of a measure matters, take $X=\{0,1\}$, $\mu ={\delta }_0$, and $K(1,1)=1$ with all other values zero. This continuous positive-semidefinite kernel gives the zero integral operator on $L^2(\mu )$, so its eigenfunctions cannot recover the value at the unmeasured point1. For a full-support measure, integration of the diagonal expansion instead gives ${\sum }_j{\lambda }_j={\int }_{X}K(x,x)d\mu (x)$. This connects the trace-class operator trace to the diagonal of its continuous integral kernel.

The strict distance inequality also gives $d(fx,fy)\le d(x,y)$ when $x=y$, so $f$ is Lipschitz continuous with constant at most one, hence continuous. Put $g(x)=d(x,f(x))$. The triangle inequality givesThus $g$ is also Lipschitz continuous and continuous.

Assume the compact metric space is nonempty, as needed for existence of a point. By the extreme value theorem, $g$ attains its infimum at some $p$. If $h=g(p)>0$, then $p\ne f(p)$, and strict distance decrease givescontradicting minimality. Since distances are nonnegative, $h=0$ and $f(p)=p$. If two different fixed points $p,q$ existed, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Therefore there is exactly one fixed point. This is strict distance decrease on a compact metric space; no uniform contraction factor less than one was assumed.

The strict distance inequality implies $d(fx,fy)\le d(x,y)$ for all points, so $f$ is Lipschitz continuous. Henceis a continuous function, since $|g(x)-g(y)|\le d(fx,fy)+d(x,y)\le 2d(x,y)$. On the nonempty compact metric space, $g$ attains its minimum at some $p$.

If $fp\ne p$, the strict distance inequality applied to $fp,p$ givesa contradiction. Thus $fp=p$. If $p,q$ were different fixed points, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. There is exactly one fixed point. This proves strict distance decrease on a compact metric space without assuming a uniform contraction constant.

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.