[proofplan] We prove independence by exhibiting one example for each direction of failure. For minimality without unique ergodicity, we cite Furstenberg's 1961 construction of a minimal skew-product homeomorphism on the two-torus admitting two distinct invariant Borel probability measures. For unique ergodicity without minimality, we construct from scratch a compact countable subset of $\mathbb{R}$ on which a continuous map $T$ pushes every non-fixed point one step toward the fixed point $0$; we then verify by direct measure computation that the Dirac mass $\delta_0$ is the only $T$-invariant Borel probability measure, while $\{0\}$ itself is a non-empty proper closed $T$-invariant subset, so $(X,T)$ is not minimal. [/proofplan] [step:Cite Furstenberg's minimal skew product with two invariant measures] Let $\mathbb{T} := \mathbb{R}/\mathbb{Z}$ be the additive circle group equipped with its quotient topology, and let $\mathbb{T}^2 := \mathbb{T} \times \mathbb{T}$ carry the product topology. By Furstenberg's construction (H. Furstenberg, *Strict ergodicity and transformation of the torus*, Amer. J. Math. **83** (1961), 573–601), there exist an irrational number $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ and a continuous map \begin{align*} f: \mathbb{T} &\to \mathbb{T} \end{align*} such that the skew product \begin{align*} S: \mathbb{T}^2 &\to \mathbb{T}^2 \\ (x, y) &\mapsto (x + \alpha,\; y + f(x)) \end{align*} is a homeomorphism, is minimal as a topological dynamical system, and admits two distinct $S$-invariant Borel probability measures $\mu_1, \mu_2$ on $\mathbb{T}^2$. We verify that $(\mathbb{T}^2, S)$ falls in the category of the theorem statement: $\mathbb{T}^2$ is non-empty, compact (as a continuous quotient of $[0,1]^2$), and metrizable (with the quotient metric induced from $\mathbb{R}^2$), and $S$ is continuous (a homeomorphism, in fact). Minimality of $(\mathbb{T}^2, S)$ means every $S$-orbit is dense in $\mathbb{T}^2$. The existence of two distinct $S$-invariant Borel probability measures $\mu_1 \neq \mu_2$ contradicts unique ergodicity, the definition of which requires the $S$-invariant Borel probability measure on $\mathbb{T}^2$ to be unique. Hence $(\mathbb{T}^2, S)$ is a minimal topological dynamical system that is not uniquely ergodic, establishing part (i). [guided] For part (i) we need an example whose every orbit is dense in the phase space but whose space of invariant Borel probability measures contains more than one element. The classical example is due to Furstenberg. Let $\mathbb{T} := \mathbb{R}/\mathbb{Z}$ be the circle, equipped with its additive group structure and the quotient topology coming from $\mathbb{R}$, and let $\mathbb{T}^2 := \mathbb{T} \times \mathbb{T}$ with the product topology. The construction of H. Furstenberg, *Strict ergodicity and transformation of the torus*, Amer. J. Math. **83** (1961), 573–601, produces an explicit irrational number $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ and an explicit continuous function \begin{align*} f: \mathbb{T} &\to \mathbb{T} \end{align*} for which the skew product \begin{align*} S: \mathbb{T}^2 &\to \mathbb{T}^2 \\ (x, y) &\mapsto (x + \alpha,\; y + f(x)) \end{align*} is a homeomorphism, is minimal, and admits two distinct $S$-invariant Borel probability measures $\mu_1, \mu_2$. Before using this in the theorem, we check that $(\mathbb{T}^2, S)$ meets the standing hypotheses on $(X, T)$: $\mathbb{T}^2$ is non-empty; it is compact, being the continuous image of the compact square $[0,1]^2$ under the quotient map $\mathbb{R}^2 \to \mathbb{T}^2$; it is metrizable, since the quotient metric $d([x],[y]) := \inf_{n,m \in \mathbb{Z}} \sqrt{(x_1 - y_1 - n)^2 + (x_2 - y_2 - m)^2}$ induces the quotient topology; and $S$ is continuous (every coordinate of $S$ is continuous on $\mathbb{T}^2$). Hence $(\mathbb{T}^2, S)$ is a topological dynamical system in the sense of the theorem statement, and in fact $S$ is a homeomorphism. Minimality of $(\mathbb{T}^2, S)$ means $\overline{\{S^n(p) : n \in \mathbb{Z}\}} = \mathbb{T}^2$ for every $p \in \mathbb{T}^2$. Unique ergodicity would require that there is exactly one $S$-invariant Borel probability measure on $\mathbb{T}^2$; the existence of two distinct such measures $\mu_1 \neq \mu_2$ contradicts this requirement. Hence $(\mathbb{T}^2, S)$ is minimal but not uniquely ergodic, which is the assertion of part (i). [/guided] [/step] [step:Construct a compact countable system attracted to a fixed point] Define the subset \begin{align*} X := \{0\} \cup \left\{ \tfrac{1}{n} : n \in \mathbb{N},\; n \geq 1 \right\} \subseteq \mathbb{R}, \end{align*} equipped with the subspace topology inherited from the standard topology on $\mathbb{R}$. Define the map \begin{align*} T: X &\to X \\ 0 &\mapsto 0, \\ \tfrac{1}{n} &\mapsto \tfrac{1}{n+1} \qquad (n \in \mathbb{N},\; n \geq 1). \end{align*} We verify that $(X, T)$ is a topological dynamical system in the sense of the theorem statement: $X$ is non-empty (it contains $0$); $X \subseteq [0,1]$ is bounded, and it is closed in $\mathbb{R}$ because its only accumulation point in $\mathbb{R}$ is $0$, which lies in $X$, so by the Heine–Borel theorem $X$ is compact; $X$ is metrizable as a subspace of the metric space $\mathbb{R}$. The map $T$ is continuous. Every point $1/n$ is isolated in $X$ (the open interval $(1/(n+1),\, 1/(n-1)) \cap X$ equals $\{1/n\}$, with the convention $1/0 = +\infty$ when $n = 1$), and at an isolated point every function is continuous. The only non-isolated point of $X$ is $0$. To verify continuity at $0$ we use the sequential characterization of continuity in the metric space $X$: if $(x_k)_{k \geq 1} \subseteq X$ satisfies $x_k \to 0$, then either $x_k = 0$ (giving $T(x_k) = 0$) or $x_k = 1/n_k$ with $n_k \to \infty$ (giving $T(x_k) = 1/(n_k + 1) \to 0$); in either case $T(x_k) \to 0 = T(0)$. Hence $T: X \to X$ is continuous and $(X, T)$ is a topological dynamical system. [guided] We now construct the second example explicitly so that every hypothesis is verifiable from the formulas. Let \begin{align*} X := \{0\} \cup \left\{ \tfrac{1}{n} : n \in \mathbb{N},\; n \geq 1 \right\} \subseteq \mathbb{R} \end{align*} with the subspace topology inherited from $\mathbb{R}$. Geometrically, $X$ is the sequence $1,\, 1/2,\, 1/3,\, \dots$ together with its limit point $0$. The theorem statement requires that the phase space be non-empty, compact, and metrizable. We check each property explicitly. - *Non-empty*: $0 \in X$. - *Compact*: By the Heine–Borel theorem, a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded. We have $X \subseteq [0,1]$, so $X$ is bounded. For closedness in $\mathbb{R}$: the only accumulation point of the set $\{1/n : n \geq 1\}$ in $\mathbb{R}$ is $0$, and $0 \in X$, so $X$ contains all of its accumulation points and is therefore closed. By Heine–Borel, $X$ is compact. - *Metrizable*: $X$ inherits the metric $d(x, y) := |x - y|$ from $\mathbb{R}$, and this metric induces the subspace topology on $X$. Define the map \begin{align*} T: X &\to X \\ 0 &\mapsto 0, \\ \tfrac{1}{n} &\mapsto \tfrac{1}{n+1} \qquad (n \in \mathbb{N},\; n \geq 1). \end{align*} We verify that $T$ is continuous. At each point $1/n$ with $n \geq 1$, the singleton $\{1/n\}$ is open in $X$: the open interval \begin{align*} \left( \tfrac{1}{n+1},\; \tfrac{1}{n-1} \right) \cap X = \{ 1/n \} \end{align*} (with the convention $1/0 = +\infty$ when $n = 1$) is open in $X$ and contains only $1/n$. So $1/n$ is an isolated point of $X$, and every function on $X$ is continuous at every isolated point. The only non-isolated point of $X$ is $0$. To check continuity of $T$ at $0$, we use the sequential characterization of continuity in metric spaces: $T$ is continuous at $0$ if and only if $T(x_k) \to T(0) = 0$ for every sequence $(x_k)_{k \geq 1} \subseteq X$ with $x_k \to 0$. Take any such sequence. If $x_k = 0$, then $T(x_k) = 0$. If $x_k = 1/n_k$ for some $n_k \geq 1$, then $1/n_k \to 0$ forces $n_k \to \infty$, so \begin{align*} T(x_k) = \tfrac{1}{n_k + 1} \to 0. \end{align*} In both cases $T(x_k) \to 0 = T(0)$, which proves $T$ is continuous at $0$. Hence $T: X \to X$ is continuous and $(X, T)$ is a topological dynamical system on a non-empty compact metrizable space, as the theorem requires. [/guided] [/step] [step:Prove that the Dirac mass at the fixed point is the unique invariant Borel probability measure] Let $\mu$ be any $T$-invariant Borel probability measure on $X$. For each $n \in \mathbb{N}$ with $n \geq 1$, define the Borel singleton \begin{align*} A_n := \{ 1/n \} \subseteq X. \end{align*} (Singletons are closed in the metric space $X$, hence Borel.) From the definition of $T$, the preimages are \begin{align*} T^{-1}(A_1) = \varnothing, \qquad T^{-1}(A_n) = A_{n-1} \quad (n \geq 2), \end{align*} since no point of $X$ is sent to $1$, and for $n \geq 2$ the unique preimage of $1/n$ is $1/(n-1)$. $T$-invariance of $\mu$ means $\mu(B) = \mu(T^{-1}(B))$ for every Borel $B \subseteq X$. Applying this with $B = A_1$, \begin{align*} \mu(A_1) = \mu(T^{-1}(A_1)) = \mu(\varnothing) = 0. \end{align*} For $n \geq 2$, the same identity gives \begin{align*} \mu(A_n) = \mu(T^{-1}(A_n)) = \mu(A_{n-1}), \end{align*} so induction on $n$ yields $\mu(A_n) = 0$ for every $n \geq 1$. The non-zero points of $X$ form a countable union of these singletons, so by countable subadditivity of $\mu$, \begin{align*} \mu\bigl( \{ 1/n : n \geq 1 \} \bigr) \leq \sum_{n=1}^{\infty} \mu(A_n) = 0. \end{align*} Since $X = \{0\} \sqcup \{1/n : n \geq 1\}$ and $\mu(X) = 1$, we conclude \begin{align*} \mu(\{0\}) = 1. \end{align*} A Borel probability measure on $X$ giving mass $1$ to $\{0\}$ is the Dirac mass $\delta_0$ at $0$, defined by $\delta_0(B) := 1$ if $0 \in B$ and $\delta_0(B) := 0$ otherwise. Hence $\mu = \delta_0$. Conversely, $\delta_0$ is $T$-invariant: since $T(0) = 0$, for every Borel $B \subseteq X$ we have $0 \in T^{-1}(B) \iff T(0) \in B \iff 0 \in B$, so $\delta_0(T^{-1}(B)) = \delta_0(B)$. Therefore $\delta_0$ is the unique $T$-invariant Borel probability measure on $X$, and $(X, T)$ is uniquely ergodic. [guided] Let $\mu$ be an arbitrary $T$-invariant Borel probability measure on $X$. The strategy is to prove that $\mu$ must place all of its mass at the fixed point $0$, so that $\mu$ is forced to equal the Dirac mass $\delta_0$. For each $n \in \mathbb{N}$ with $n \geq 1$, define \begin{align*} A_n := \{ 1/n \} \subseteq X. \end{align*} Each $A_n$ is Borel: in the metric space $X$ every singleton is closed, and closed sets are Borel. We compute the preimages $T^{-1}(A_n)$. The equation $T(x) = 1/n$ has the unique solution $x = 1/(n-1)$ when $n \geq 2$, and no solution when $n = 1$ (the values $T$ takes on the non-zero points $1/k$ are $1/(k+1) \in \{1/2,\, 1/3,\, \dots\}$, never $1$, and $T(0) = 0$). Hence \begin{align*} T^{-1}(A_1) = \varnothing, \qquad T^{-1}(A_n) = A_{n-1} \quad (n \geq 2). \end{align*} $T$-invariance of $\mu$ asserts that $\mu(B) = \mu(T^{-1}(B))$ for every Borel $B \subseteq X$. Applying invariance with $B = A_1$, \begin{align*} \mu(A_1) = \mu(T^{-1}(A_1)) = \mu(\varnothing) = 0. \end{align*} Applying invariance again with $B = A_n$ for $n \geq 2$, \begin{align*} \mu(A_n) = \mu(T^{-1}(A_n)) = \mu(A_{n-1}). \end{align*} By induction on $n$, starting from $\mu(A_1) = 0$, we obtain $\mu(A_n) = 0$ for every $n \geq 1$. The non-zero points of $X$ form a countable union of singletons: $\{ 1/n : n \geq 1 \} = \bigcup_{n=1}^{\infty} A_n$. Countable subadditivity of the probability measure $\mu$ gives \begin{align*} \mu\bigl( \{ 1/n : n \geq 1 \} \bigr) \leq \sum_{n=1}^{\infty} \mu(A_n) = 0. \end{align*} Since $X = \{0\} \sqcup \{ 1/n : n \geq 1 \}$ is a disjoint decomposition and $\mu$ is a probability measure, \begin{align*} \mu(\{0\}) = \mu(X) - \mu\bigl( \{ 1/n : n \geq 1 \} \bigr) = 1 - 0 = 1. \end{align*} A Borel probability measure on $X$ that gives mass $1$ to the singleton $\{0\}$ is the Dirac mass $\delta_0$, defined by $\delta_0(B) = 1$ if $0 \in B$ and $\delta_0(B) = 0$ otherwise. Hence $\mu = \delta_0$. Conversely, we verify that $\delta_0$ itself is $T$-invariant. For every Borel $B \subseteq X$, the membership $0 \in T^{-1}(B)$ is equivalent to $T(0) \in B$, and since $T(0) = 0$ this is equivalent to $0 \in B$. Therefore $\delta_0(T^{-1}(B)) = \delta_0(B)$ for every Borel $B$, which is the definition of $T$-invariance. We have shown two things: $\delta_0$ is $T$-invariant, and every $T$-invariant Borel probability measure equals $\delta_0$. Together these say that $\delta_0$ is the unique $T$-invariant Borel probability measure on $X$, which is exactly the definition of $(X, T)$ being uniquely ergodic. [/guided] [/step] [step:Exhibit a non-empty proper closed invariant subset and conclude] Define \begin{align*} M := \{ 0 \} \subseteq X. \end{align*} We verify that $M$ is non-empty (it contains $0$), closed (singletons are closed in the metric space $X$, which is in particular Hausdorff), proper (e.g., $1 \in X \setminus M$), and $T$-invariant (since $T(0) = 0$, we have $T(M) = \{0\} \subseteq M$). We invoke the standard equivalence: a topological dynamical system $(Y, S)$ with $Y$ a non-empty compact Hausdorff space is minimal if and only if $Y$ contains no non-empty proper closed $S$-invariant subset (see, e.g., P. Walters, *An Introduction to Ergodic Theory*, GTM 79, Springer, 1982, Theorem 5.9). Since $M = \{0\}$ is a non-empty proper closed $T$-invariant subset of $X$, $(X, T)$ is not minimal. Combined with the previous step, $(X, T)$ is a uniquely ergodic topological dynamical system that is not minimal, establishing part (ii). Combining part (i) — Furstenberg's homeomorphism $(\mathbb{T}^2, S)$, which is minimal but not uniquely ergodic — with part (ii) — the compact countable system $(X, T)$, which is uniquely ergodic but not minimal — we conclude that, in the category of topological dynamical systems on non-empty compact metrizable spaces with continuous dynamics, minimality and unique ergodicity are logically independent properties. This completes the proof. [guided] We now show that the system $(X, T)$ constructed above is not minimal, and then combine both examples to conclude. To prove non-minimality, we exhibit a non-empty proper closed $T$-invariant subset of $X$. Take the singleton \begin{align*} M := \{ 0 \} \subseteq X. \end{align*} We verify each of the four required properties: - *Non-empty*: $0 \in M$. - *Closed*: $X$ is a metric space and hence Hausdorff, so every singleton is closed. - *Proper*: $1 \in X$ but $1 \notin M$, so $M \subsetneq X$. - *$T$-invariant*: $T(0) = 0 \in M$, so $T(M) = \{0\} \subseteq M$. We now invoke the standard equivalence in topological dynamics: a topological dynamical system $(Y, S)$ on a non-empty compact Hausdorff space $Y$ is minimal if and only if $Y$ contains no non-empty proper closed $S$-invariant subset. The argument is short. Suppose first that every $S$-orbit is dense. If $F \subseteq Y$ is a non-empty closed $S$-invariant subset, pick any $p \in F$; then $\{S^n p : n \geq 0\} \subseteq F$ by invariance, and taking closures gives $Y = \overline{\{S^n p : n \geq 0\}} \subseteq F$, so $F = Y$, meaning $F$ is not proper. Conversely, suppose every non-empty closed $S$-invariant subset of $Y$ equals $Y$. For any $p \in Y$, the orbit closure $\overline{\{S^n p : n \geq 0\}}$ is non-empty, closed, and $S$-invariant, so it equals $Y$; that is, the orbit of $p$ is dense, so $(Y, S)$ is minimal. A textbook reference is P. Walters, *An Introduction to Ergodic Theory*, Graduate Texts in Mathematics **79** (Springer, 1982), Theorem 5.9. Since $M = \{0\}$ is a non-empty proper closed $T$-invariant subset of $X$, the system $(X, T)$ is not minimal. Combined with the previous step, where we proved that $(X, T)$ is uniquely ergodic, this gives a uniquely ergodic topological dynamical system that is not minimal, proving part (ii) of the theorem. Finally, we put the two halves of the proof together. Part (i) was established in the first step by citing Furstenberg's skew-product homeomorphism $(\mathbb{T}^2, S)$, which is minimal but admits two distinct invariant Borel probability measures and hence is not uniquely ergodic. Part (ii) has just been established with the explicit system $(X, T)$, which is uniquely ergodic with unique invariant measure $\delta_0$ but admits the non-empty proper closed invariant subset $\{0\}$ and hence is not minimal. Together these examples show that, in the category of topological dynamical systems on non-empty compact metrizable spaces with continuous dynamics, neither of the two properties — minimality and unique ergodicity — implies the other. This is precisely the statement that minimality and unique ergodicity are logically independent, and the proof is complete. [/guided] [/step]