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A nonempty compact dynamical system $(X,T)$ is a minimal dynamical system if it has no proper nonempty closed forward-invariant subset. Equivalently, every forward orbit is dense: the orbit closure of any point is a nonempty closed forward-invariant set, and conversely a closed invariant set contains the orbit closure of each of its points.

Consider the collection of nonempty closed forward-invariant subsets of $X$, ordered by reverse inclusion. Every chain has an upper bound, its intersection: the chain has the finite-intersection property, compactness makes the intersection nonempty, and continuity is not even needed for its forward invariance. The Zorn lemma supplies a minimal member $Y$, so $(Y,T{|}_Y)$ is a minimal subsystem. Also $T(Y)$ is compact, nonempty and forward invariant, so minimality gives $T(Y)=Y$. Here the printed word “nontrivial” must mean nonempty; a singleton with its identity map shows that a subsystem with more than one point cannot always be required.

If $y\in Y$, every tail of its forward orbit is dense in $Y$, since it is the orbit of $T^{N}y\in Y$. Therefore every neighborhood of $y$ contains $T^{n}y$ for arbitrarily large $n$. Choosing these times successively for balls of radii $1/j$ gives $n_j\to \infty$ with $T^{n_j}y\to y$. This proves the Birkhoff recurrence theorem: every continuous map of a nonempty compact metric space has a recurrent point. In fact every point of the chosen minimal subsystem is recurrent.

Write the continuous circle extension as the skew productA sufficient condition, and in this setting also a necessary one, is the circle skew-product minimality criterion: for every integer $m\ne 0$ there is no continuous function $f:X\to {\mathbb{S}}^1$ such thatThe cocycle $\rho$ is assumed continuous; a merely measurable cocycle need not define this topological extension.

To prove sufficiency, choose a minimal subsystem $Z$ of the extension. Its projection is a nonempty closed forward-invariant subset of $X$, hence all of $X$. For $a\in \mathbb{R}/\mathbb{Z}$, vertical translation sends $Z$ to another minimal subsystem, since it commutes with $\tilde{T}$. Two such translates either are disjoint or coincide: a nonempty intersection is closed and forward invariant, so minimality forces equality. Define their common stabilizerThis is a closed subgroup of the circle group. Its closedness follows by taking limits of translations of points in the compact set $Z$, using both a translation and its inverse. Each fiber $Z_x$ is nonempty. If $y,z\in Z_x$, the sets $Z$ and $Z+(z-y)$ intersect at $(x,z)$, so $z-y\in H$. Conversely translating any fiber point by any member of $H$ stays in the fiber. Thus every fiber is exactly one coset of $H$.

If $Z$ is proper, $H$ is proper and therefore finite, say $H=\{0,1/m,\dots ,(m-1)/m\}$ for some positive integer $m$. The character $y\mapsto e^{2\pi imy}$ is constant on each fiber coset, giving a well-defined function $f(x)$ from $Z_x$. This function is continuous: its graph is the continuous image of $Z$ in $X\times {\mathbb{S}}^1$, and projection of that compact graph onto the Hausdorff space $X$ is a continuous bijection and hence a homeomorphism. The relation $\tilde{T}(Z)=Z$ gives $f(Tx)=e^{2\pi im\rho (x)}f(x)$, contrary to the criterion. Hence $Z=\tilde{X}$ and the extension is minimal.

Conversely, a solution of the displayed equation makes $F(x,y)=e^{2\pi imy}/f(x)$ a nonconstant continuous invariant function on the extension. Its nonempty level sets are proper closed invariant subsets, contradicting minimality. This also proves necessity of the circle skew-product minimality criterion.

Apply it with $\alpha =\sqrt{2}$, $X=\mathbb{R}/\mathbb{Z}$, $Tx=x+\alpha$, and $\rho (x)=2x+\alpha$. The base is a minimal dynamical system: the closure of the cyclic subgroup generated by an irrational angle is the whole circle group, and its positive forward orbit has the same closure. To see the subgroup assertion, a proper closed subgroup of the circle is finite; if it contained an irrational angle it would be infinite. The elementary classification follows because a closed subgroup with arbitrarily small nonzero elements is dense, while otherwise its least positive representative generates a finite subgroup.

There can be no forbidden $f$ for any $m\ne 0$. Indeed, the winding number of a continuous closed path of $x\mapsto f(x+\alpha )$ equals that of $x\mapsto f(x)$, by translation and homotopy invariance of winding number. Consequently their quotient has winding number zero. But the proposed quotient $e^{2\pi im(2x+\alpha )}$ has winding number $2m\ne 0$. Thus the skew product is minimal. Starting at $(0,0)$, induction givesbecause $n^2\alpha +2n\alpha +\alpha =(n+1{)}^2\alpha$. Its orbit is dense in the two-dimensional torus, so its second coordinate is dense in the circle. Removing the initial point does not change density, since every orbit tail in a minimal dynamical system is dense. Therefore