We now begin the study of recurrence in topological dynamical systems $(X, {\mathcal F}, T)$ – how often a non-empty open set U in X returns to intersect itself, or how often a point x in X returns to be close to itself. Not every set or point needs to return to itself; consider for instance what happens to the shift $x \mapsto x+1$ on the compactified integers $\{-\infty\} \cup {\Bbb Z} \cup \{+\infty\}$. Nevertheless, we can always show that at least one set (from any open cover) returns to itself:

Theorem 1. (Simple recurrence in open covers) Let $(X,{\mathcal F},T)$ be a topological dynamical system, and let $(U_\alpha)_{\alpha \in A}$ be an open cover of X. Then there exists an open set $U_\alpha$ in this cover such that $U_\alpha \cap T^n U_\alpha \neq \emptyset$ for infinitely many n.

Proof. By compactness of X, we can refine the open cover to a finite subcover. Now consider an orbit $T^{\Bbb Z} x = \{ T^n x: n \in {\Bbb Z} \}$ of some arbitrarily chosen point $x \in X$. By the infinite pigeonhole principle, one of the sets $U_\alpha$ must contain an infinite number of the points $T^n x$ counting multiplicity; in other words, the recurrence set $S := \{ n: T^n x \in U_\alpha \}$ is infinite. Letting $n_0$ be an arbitrary element of S, we thus conclude that $U_\alpha \cap T^{n_0-n} U_\alpha$ contains $T^{n_0} x$ for every $n \in S$, and the claim follows. $\Box$

Exercise 1. Conversely, use Theorem 1 to deduce the infinite pigeonhole principle (i.e. that whenever ${\Bbb Z}$ is coloured into finitely many colours, one of the colour classes is infinite). Hint: look at the orbit closure of c inside $A^{\Bbb Z}$, where A is the set of colours and $c: {\Bbb Z} \to A$ is the colouring function.) $\diamond$

Now we turn from recurrence of sets to recurrence of individual points, which is a somewhat more difficult, and highlights the role of minimal dynamical systems (as introduced in the previous lecture) in the theory. We will approach the subject from two (largely equivalent) approaches, the first one being the more traditional “epsilon and delta” approach, and the second using the Stone-Čech compactification $\beta {\Bbb Z}$ of the integers (i.e. ultrafilters).

Before we begin, it will be notationally convenient to place a metric d on our compact metrisable space X [though, as an exercise, the reader is encouraged to recast all the material here in a manner which does not explicitly mention a metric]. There are of course infinitely many metrics that one could place here, but they are all coarsely equivalent in the following sense: if d, d’ are two metrics on X, then for every $\delta > 0$ there exists an $\varepsilon > 0$ such that $d'(x,y) < \delta$ whenever $d(x,y) < \varepsilon$, and similarly with the role d and d’ reversed. This claim follows from the standard fact that continuous functions between compact metric spaces are uniformly continuous. Because of this equivalence, it will not actually matter for any of our results what metric we place on our spaces. For instance, we could endow a Bernoulli system $A^{\Bbb Z}$, where A is itself a compact metrisable space (and thus $A^{\Bbb Z}$ is compact by Tychonoff’s theorem), with the metric

$d( (a_n)_{n \in {\Bbb Z}}, (b_n)_{n \in {\Bbb Z}} ) := \sum_{n \in {\Bbb Z}} 2^{-|n|} d_A(a_n,b_n)$ (1)

where $d_A$ is some arbitrarily selected metric on A. Note that this metric is not shift-invariant.

Exercise 2. Show that if A contains at least two points, then the Bernoulli system $A^{\Bbb Z}$ (with the standard shift) cannot be endowed with a shift-invariant metric. (Hint: find two distinct points which converge to each other under the shift map.) $\diamond$

Fix a metric d. For each n, the shift $T^n: X \to X$ is continuous, and hence uniformly continuous since X is compact, thus for every $\delta > 0$ there exists $\varepsilon > 0$ depending on $\delta$ and n such that $d( T^n x, T^n y ) < \delta$ whenever $d(x,y) < \varepsilon$. However, we caution that the $T^n$ need not be uniformly equicontinuous; the quantity $\varepsilon$ appearing above can certainly depend on n. Indeed, they need not even be equicontinuous. For instance, this will be the case for the Bernoulli shift with the metric (1) (why?), and more generally for any system that exhibits “mixing” or other chaotic behaviour. At the other extreme, in the case of isometric systems – systems in which T preserves the metric d – the shifts $T^n$ are all isometries, and thus are clearly uniformly equicontinuous.

We can now classify points x in X based on the dynamics of the orbit $T^{\Bbb Z} x := \{ T^n x: n \in {\Bbb Z} \}$:

1. x is invariant if $Tx=x$.
2. x is periodic if $T^n x = x$ for some non-zero n.
3. x is almost periodic if for every $\varepsilon > 0$, the set $\{ n \in {\Bbb Z}: d(T^n x, x) < \varepsilon \}$ is syndetic (i.e. it has bounded gaps);
4. x is recurrent if for every $\varepsilon > 0$, the set $\{ n \in {\Bbb Z}: d(T^n x, x) < \varepsilon \}$ is infinite. Equivalently, there exists a sequence $n_j$ of integers with $|n_j| \to \infty$ such that $\lim_{j \to \infty} T^{n_j} x = x$.

It is clear that every invariant point is periodic, that every periodic point is almost periodic, and every almost periodic point is recurrent. These inclusions are all strict. For instance, in the circle shift system $({\Bbb R}/{\Bbb Z}, x \mapsto x + \alpha)$ with $\alpha \in {\Bbb R}$ irrational, it turns out that every point is almost periodic, but no point is periodic.

Exercise 3. In the boolean Bernoulli system $(2^{\Bbb Z}, A \mapsto A-1)$, show that the discrete Cantor set

$x := \bigcup_{N=1}^\infty \{ \sum_{n=0}^N \epsilon_n 10^n: \epsilon_n \in \{-1,0,+1\} \}$ (2)

is recurrent but not almost periodic. $\diamond$

In a general topological dynamical system, it is quite possible to have points which are non-recurrent (as the example of the compactified integer shift already shows). But if we restrict to a minimal dynamical system, things get much better:

Lemma 1. If $(X,{\mathcal F},T)$ is a minimal topological dynamical system, then every element of X is almost periodic (and hence recurrent).

Proof. Suppose for contradiction that we can find a point x of X which is not almost periodic. This means that we can find $\varepsilon > 0$ such that the set $\{ n: d(T^n x, x) < \varepsilon \}$ is not syndetic. Thus, for any $m > 0$, we can find an $n_m$ such that $d(T^n x,x) \geq \varepsilon$ for all $n \in [n_m-m,n_m+m]$ (say).

Since X is compact, the sequence $T^{n_m} x$ must have at least one limit point y. But then one verifies (using the continuity of the shift operators) that

$d( T^h y, x ) = \lim_{m \to \infty} d( T^{n_m+h} x, x ) \geq \varepsilon$ (3)

for all h. But this means that the orbit closure $\overline{T^{\Bbb Z} y}$ of y does not contain x, contradicting the minimality of X. The claim follows. $\Box$

Exercise 4. If x is a point in a topological dynamical system, show that x is almost periodic if and only if it lies in a minimal system. Because of this, almost periodic points are sometimes referred to as minimal points. $\diamond$

Combining Lemma 1 with Lemma 1 of the previous lecture, we immediately obtain the

Birkhoff recurrence theorem. Every topological dynamical system contains at least one point x which is almost periodic (and hence recurrent).

Note that this is stronger than Theorem 1, as can be seen by considering the element $U_\alpha$ of the open cover which contains the almost periodic point. Indeed, we now have obtained a stronger conclusion, namely that the set of return times $\{ n: T^n U_\alpha \cap U_\alpha \neq \emptyset \}$ is not only infinite, it is syndetic.

Exercise 5. State and prove a version of the Birkhoff recurrence theorem in which the map $T: X \to X$ is continuous but not assumed to be invertible. (Of course, all references to ${\Bbb Z}$ now need to be replaced with ${\Bbb N}$.) $\diamond$

The Birkhoff recurrence theorem does not seem particularly strong, as it only guarantees existence of a single recurrent (or almost periodic point). For general systems, this is inevitable, because it can happen that the majority of the points are non-recurrent (look at the compactified integer shift system, for instance). However, suppose the system is a group quotient $(G/\Gamma, x \mapsto gx)$. To make this a topological dynamical system, we need G to be a topological group, and $\Gamma$ to be a cocompact subgroup of G (such groups are also sometimes referred to as uniform subgroups). Then we see that the system is a homogeneous space: given any two points $x, y \in G/\Gamma$, there exists a group element $h \in G$ such that hx=y. Thus we expect any two points in $G/\Gamma$ to behave similarly to each other. Unfortunately, this does not quite work in general, because the action of h need not preserve the shift $x \mapsto gx$, as there is no reason that h commutes with g. But suppose that g is a central element of G (which is for instance the case if G is abelian). Then the action of h is now an isomorphism on the dynamical system $(G/\Gamma, x \mapsto gx)$. In particular, if hx = y, we see that x is almost periodic (or recurrent) if and only if y is. We thus conclude

Theorem 2. (Kronecker type approximation theorem) Let $(G/\Gamma, x \mapsto gx)$ be a topological group quotient dynamical system such that g lies in the centre Z(G) of G. Then every point in this system is almost periodic (and hence recurrent).

Applying this theorem to the torus shift $(({\Bbb R}/{\Bbb Z})^d, x \mapsto x+\alpha)$, where $\alpha = (\alpha_1,\ldots,\alpha_d) \in {\Bbb R}^d$ is a vector, we thus obtain that for any $\varepsilon > 0$, the set

$\{ n \in {\Bbb Z}: \hbox{dist}( n \alpha, {\Bbb Z}^d ) < \varepsilon \}$ (4)

is syndetic (and in particular, infinite). This should be compared with the classical Kronecker approximation theorem.

It is natural to ask what happens when g is not central. If G is a Lie group and the action of g on the Lie algebra ${\mathfrak g}$ is unipotent rather than trivial, then Theorem 2 still holds; this follows from Ratner’s theorem, of which we will discuss much later in this course. But the claim is not true for all group quotients. Consider for instance the Bernoulli shift system $(X,T) = ( ({\Bbb Z}/2{\Bbb Z})^{\Bbb Z}, T)$, which is isomorphic to the boolean Bernoulli shift system. As the previous examples have already shown, this system contains both recurrent and non-recurrent elements. On the other hand, it is intuitive that this system has a lot of symmetry, and indeed we can view it as a group quotient $(G/\Gamma, x \mapsto gx)$. Specifically, G is the lamplighter group $G = {\Bbb Z}/2{\Bbb Z} \wr {\Bbb Z}$. To describe this group, we observe that the group $({\Bbb Z}/2{\Bbb Z})^{\Bbb Z}$ acts on X by addition, whilst the group ${\Bbb Z}$ acts on X via the shift map T. The lamplighter group $G := ({\Bbb Z}/2{\Bbb Z})^{\Bbb Z} \times {\Bbb Z}$ then acts by both addition and shift:

$(a,n): x \mapsto T^n x + a$ for all $(a,n) \in G$. (5)

In order for this to be a group action, we endow G with the multiplication law

$(a,n) (b,m) := (a + T^n b, n+m)$; (6)

one easily verifies that this really does make G into a group; if we give G the product topology, it is a topological group. G clearly acts transitively on the compact space X, and so $X \equiv G/\Gamma$ for some cocompact subgroup $\Gamma$ (which turns out to be isomorphic to ${\Bbb Z}$ – why?). By construction, the shift map T can be expressed using the group element $(0,1) \in G$, and so we have turned the Bernoulli system into a group quotient. Since this system contains non-recurrent points (e.g. the indicator function of the natural numbers) we see that Theorem 2 does not hold for arbitrary group quotients.

– The ultrafilter approach –

Now we turn to a different approach to topological recurrence, which relies on compactifying the underlying group ${\Bbb Z}$ that acts on topological dynamical systems. By doing so, all the epsilon management issues go away, and the subject becomes very algebraic in nature. On the other hand, some subtleties arise also; for instance, the compactified object $\beta {\Bbb Z}$ is not a group, but merely a left-continuous semigroup.

This approach is based on ultrafilters or (equivalently) via the Stone-Čech compactification. Let us recall how this compactification works:

Theorem 3. (Stone-Čech compactification) Every locally compact Hausdorff (LCH) space X can be embedded in a compact Hausdorff space $\beta X$ in which X is an open dense set. (In particular, if X is already compact, then $\beta X = X$.) Furthermore, any continuous function $f: X \to Y$ between LCH spaces extends uniquely to a continuous function $\beta f: \beta X \to \beta Y$.

Proof. (Sketch) This proof uses the intuition that $\beta X$ should be the “finest” compactification of X. Recall that a compactification of a LCH space X is any compact Hausdorff space containing X as an open dense set. We say that one compactification Y of X is finer than another Z if there is a surjective continuous map from Y to Z that is the identity on X. (Note that as X is dense in Y, and Z is Hausdorff, this surjection is unique.) For instance, the two-point compactification $\{-\infty\} \cup {\Bbb Z} \cup \{+\infty\}$ of the integers is finer than the one-point compactification ${\Bbb Z} \cup \{\infty\}$. This is clearly a partial ordering; also, the inverse limit of any chain of compactifications can be verified (by Tychonoff’s theorem) to still be a compactification. Hence, by Zorn’s lemma (modulo a technical step in which one shows that the moduli space of compactifications of X is a set rather than a class), there is a maximal compactification $\beta X$. To verify the extension property for continuous functions $f: X \to Y$, note (by replacing Y with $\beta Y$ if necessary) that we may take Y to be compact. Let Z be the closure of the graph $X' := \{ (x,f(x)): x \in X \}$ in $(\beta X) \times Y$. X’ is clearly homeomorphic to X, and so Z is a compactification of X. Also, there is an obvious surjective continuous map from Z to $\beta X$; thus by maximality, this map must be a homeomorphism, thus Z is the graph of a continuous function $\beta f: \beta X \to \beta Y$, and the claim follows (the uniqueness of $\beta f$ is easily established). $\Box$

Exercise 6. Let X be discrete (and thus clearly LCH), and let $\beta X$ be the Stone-Čech compactification. For any $p \in \beta X$, let ${}[p] \in 2^{2^X}$ be the collection of all sets $A \subset X$ such that $\beta 1_A(p) = 1$. Show that [p] is an ultrafilter, or in other words that it obeys the following four properties:

1. $\emptyset \not \in [p]$.
2. If $U \in [p]$ and $V \in 2^X$ are such that $U \subset V$, then $V \in [p]$.
3. If $U, V \in [p]$, then $U \cap V \in [p]$.
4. If $U, V \in 2^X$ are such that $U \cup V = X$, then at least one of U and V lie in [p].

Furthermore, show that the map $p \mapsto [p]$ is a homeomorphism between $\beta X$ and the space of ultrafilters, which we endow with the topology induced from the product topology on $2^{2^X} \equiv \{0,1\}^{2^X}$, where we give $\{0,1\}$ the discrete topology (one can place some other topologies here also). Thus we see that in the discrete case, we can represent the Stone-Čech compactification explicitly via ultrafilters. $\diamond$

It is easy to see that $\beta(g \circ f) = (\beta g) \circ (\beta f)$ whenever $f: X \to Y$ and $g: Y \to Z$ are continuous maps between LCH spaces. In the language of category theory, we thus see that $\beta$ is a covariant functor from the category of LCH spaces to the category of compact Hausdorff spaces. (The above theorem does not explicitly define $\beta X$, but it is not hard to see that this compactification is unique up to homeomorphism, so the exact form of $\beta X$ is somewhat moot. However, it is possible to create an ultrafilter-based description of $\beta X$ for general LCH spaces X, though we will not do so here.)

Exercise 6′. Let X and Y be two LCH spaces. Show that the disjoint union $(\beta X) \uplus (\beta Y)$ of $\beta X$ and $\beta Y$ is isomorphic to $\beta (X \uplus Y)$. (Indeed, this isomorphism is a natural isomorphism.) In the language of category theory, this means that $\beta$ preserves coproducts. (Unfortunately, $\beta$ does not preserve products, which leads to various subtleties, such as the non-commutativity of the compactification of commutative groups.) $\diamond$

Note that if $f: X \to Y$ is continuous, then $\beta f: \beta X \to \beta Y$ is continuous also; since X is dense in $\beta X$, we conclude that

$\beta f(p) = \lim_{x \to p} f(x)$ (7)

for all $p \in \beta X$, where x is constrained to lie in X. [Here and in the sequel, limits such as $\lim_{x \to p}$ are interpreted in the usual topological sense, thus (7) means that for every neighbourhood V of $\beta f(p)$, there exists a neighbourhood U of p such that $f(x) \in V$ for all $x \in U$.)  In particular, the limit on the right exists for any continuous $f: X \to Y$, and thus if X is discrete, it exists for any (!) function $f: X \to Y$. Each p can then be viewed as a recipe for taking limits of arbitrary functions in a consistent fashion (although different p's can give different limits, of course). It is this ability to take limits without needing to check for convergence and without running into contradictions that makes the Stone-Čech compactification a useful tool here. (See also my post on ultrafilters for further discussion.)

The integers ${\Bbb Z}$ are discrete, and thus are clearly LCH. Thus we may form the compactification $\beta {\Bbb Z}$. The addition operation $+: {\Bbb Z} \times {\Bbb Z} \to {\Bbb Z}$ can then be extended to $\beta {\Bbb Z}$ by the plausible-looking formula

$p + q := \lim_{n \to p} \lim_{m \to q} n+m$ (8)

for all $p, q \in \beta {\Bbb Z}$, where n, m range in the integers ${\Bbb Z}$. Note that the double limit is guaranteed to exist by (7). Equivalently, we have

$\lim_{l \to p+q} f(l) = \lim_{n \to p} \lim_{m \to q} f(n+m)$ (8')

for all functions $f: {\Bbb Z} \to X$ into an LCH space X; one can derive (8') from (8) by applying $\beta f: \beta {\Bbb Z} \to \beta X$ to both sides of (8) and using (7) and the continuity of $\beta f$ repeatedly.

This addition operation clearly extends that of ${\Bbb Z}$ and is associative, thus we have turned $\beta {\Bbb Z}$ into a semigroup. We caution however that this semigroup is not commutative, due to the usual difficulty that double limits in (8) cannot be exchanged. (We will prove non-commutativity shortly.) For similar reasons, $\beta {\Bbb Z}$ is not a group; the obvious attempt to define a negation operation $-p := \lim_{n \to p} -n$ is well-defined, but does not actually invert addition. The operation $(p,q) \mapsto p+q$ is continuous in p for fixed q (why?), but is not necessarily continuous in q for fixed p - again, due to the exchange of limits problem. Thus $\beta {\Bbb Z}$ is merely a left-continuous semigroup. If however p is an integer, then the first limit in (8) disappears, and one easily shows that $q \mapsto q+p$ is continuous in this case (and for similar reasons one also recovers commutativity, $q+p=p+q$).

Exercise 7. Let us endow the two-point compactification $\{-\infty\} \cup {\Bbb Z} \cup \{+\infty\}$ with the semigroup structure + in which $x + (+\infty) = +\infty$ and $x + (-\infty) = -\infty$ for all $x \in \{ -\infty\} \cup {\Bbb Z} \cup \{+\infty\}$ (compare with (8)). Show that there is a unique continuous map $\pi: \beta {\Bbb Z} \to {\Bbb Z} \cup \{-\infty\} \cup \{+\infty\}$ which is the identity on ${\Bbb Z}$, and that this map is a surjective semigroup homomorphism. Using this homomorphism, conclude:

1. $\beta{\Bbb Z}$ is not commutative. Furthermore, show that the centre $Z( \beta{\Bbb Z} ) := \{ p \in \beta{\Bbb Z}: p+q=q+p \hbox{ for all } q \in \beta{\Bbb Z} \}$ is exactly equal to ${\Bbb Z}$.
2. Show that if $p,q \in \beta {\Bbb Z}$ are such that $p+q \in {\Bbb Z}$, then $p, q \in {\Bbb Z}$. ("Once you go to infinity, you can never return.") Conclude in particular that $\beta {\Bbb Z}$ is not a group. (Note that this conclusion could already be obtained using the coarser one-point compactification ${\Bbb Z} \cup \{\infty\}$ of the integers.) $\diamond$

Remark 1. More generally, we can take any LCH left-continuous semigroup S and compactify it to obtain a compact Hausdorff left-continuous semigroup $\beta S$. Observe that if $f: S \to S'$ is a homomorphism between two LCH left-continuous semigroups, then $\beta f: \beta S \to \beta S'$ is also a homomorphism. Thus, from the viewpoint of category theory, $\beta$ can be viewed as a covariant functor from the category of LCH left-continuous semigroups to the category of CH left-continuous semigroups. We will see this functorial property being used a little later in this course. (The issue turns out to be considerably more delicate than I thought, as pointed out in comments; see this article for further discussion.) $\diamond$

The left-continuous non-commutative semigroup structure of $\beta {\Bbb Z}$ may appear to be terribly weak when compared against the jointly continuous commutative group structure of ${\Bbb Z}$, but $\beta {\Bbb Z}$ has a decisive trump card over ${\Bbb Z}$: it is compact. We will see the power of compactness a little later in this lecture.

A topological dynamical system $(X,{\mathcal F},T)$ yields an action $n \mapsto T^n$ of the integers ${\Bbb Z}$. But we can automatically extend this action to an action $p \mapsto T^p$ of the compactified integers $\beta {\Bbb Z}$ by the formula

$T^p x := \lim_{n \to p} T^n x$. (9)

(Note that X is already compact, so that the limit in (9) stays in X.) One easily checks from (8') that this is indeed an action of $\beta {\Bbb Z}$ (thus $T^p T^q = T^{p+q}$ for all $p,q \in {\Bbb Z}$). The map $T^p x$ is continuous in p by construction; however we caution that it is no longer continuous in x (it's the exchange-of-limits problem once more!). Indeed, the map $T^p: X \to X$ can be quite nasty from an analytic viewpoint; for instance, it is possible for this map to not be Borel measurable. (This is the price one pays for introducing beasts generated from the axiom of choice into one's mathematical ecosystem.) But as we shall see, the algebraic properties of $T^p$ are very good, and suffice for applications to recurrence, because once one has compactified the underlying semigroup $\beta {\Bbb Z}$, the need for point-set topology (and for all the epsilons that come with it) mostly disappears. For instance, we can now replace orbit closures by orbits:

Lemma 2. Let $(X,{\mathcal F},T)$ be a topological dynamical system, and let $x \in X$. Then

$\overline{T^{\Bbb Z}(x)} = T^{\beta {\Bbb Z}} x := \{ T^p x: p \in \beta {\Bbb Z} \}$.

Proof: Since $\beta {\Bbb Z}$ is compact, $T^{\beta {\Bbb Z}} x$ is compact also. Since ${\Bbb Z}$ is dense in $\beta {\Bbb Z}$, $T^{\Bbb Z} x$ is dense in $T^{\beta {\Bbb Z}} x$. The claim follows. $\Box$

From (9) we see that $T^p$ is some sort of "limiting shift" operation. To get some intuition, let us consider the compactified integer shift $(\{-\infty\} \cup {\Bbb Z} \cup \{+\infty\}, x \mapsto x+1)$, and look at the orbit of the point 0. If one only shifts by integers $n \in {\Bbb Z}$, then $T^n 0$ can range across the region ${\Bbb Z}$ in the system but cannot reach $-\infty$ or $+\infty$. But now let $p \in \beta {\Bbb Z} \backslash {\Bbb Z}$ be any limit point of the positive integers ${\Bbb Z}^+$ (note that at least one such limit point must exist, since ${\Bbb Z}^+$ is not compact. Indeed, in the language of Exercise 7, the set of all such limit points is $\pi^{-1}(+\infty)$.) Then from (9) we see that $T^p 0 = +\infty$. Similarly, if $q \in \beta{\Bbb Z} \backslash {\Bbb Z}$ is a limit point of the negative integers ${\Bbb Z}^-$ then $T^q 0 = -\infty$. Now, since $+\infty$ invariant, we have $T^q (+\infty) = +\infty$ by (9) again, and thus $T^q T^p 0 = +\infty$, while $T^p T^q 0 = -\infty$. In particular, we see that $p+q \neq q+p$, demonstrating non-commutativity in $\beta {\Bbb Z}$ (again, compare with Exercise 7). Informally, the problem here is that in (8), n+m will go to $+\infty$ if we let m go to $+\infty$ first and then $n \to -\infty$ next, but if we take $n \to -\infty$ first and then $m \to +\infty$ next, n+m instead goes to $-\infty$.

Exercise 8. Let $A \subset {\Bbb Z}$ be a set of integers.

1. Show that $\beta A$ can be canonically identified with the closure of A in $\beta {\Bbb Z}$, in which case $\beta A$ becomes a clopen subset of $\beta {\Bbb Z}$.
2. Show that A is infinite if and only if $\beta A \not \subset {\Bbb Z}$.
3. Show that A is syndetic if and only if $\beta A \cap (\beta{\Bbb Z} + p) \neq \emptyset$ for every $p \in \beta {\Bbb Z}$. (Since $\beta A$ is clopen, this condition is also equivalent to requiring $\beta A \cap ({\Bbb Z} + p) \neq \emptyset$ for every $p \in \beta {\Bbb Z}$.)
4. A set of integers A is said to be thick if it contains arbitrarily long intervals [a_n,a_n+n]; thus syndetic and thick sets always intersect each other. Show that A is thick if and only if there exists $p \in \beta{\Bbb Z}$ such that $\beta{\Bbb Z}+p \subset \beta A$. (Again, this condition is equivalent to requiring ${\Bbb Z}+p \subset \beta A$ for some p.) $\diamond$

Recall that a system is minimal if and only if it is the orbit closure of every point in that system. We thus have a purely algebraic description of minimality:

Corollary 1. Let $(X,{\mathcal F},T)$ be a topological dynamical system. Then X is minimal if and only if the action of $\beta {\Bbb Z}$ is transitive; thus for every $x, y \in {\Bbb Z}$ there exists $p \in \beta {\Bbb Z}$ such that $T^p x = y$.

One also has purely algebraic descriptions of almost periodicity and recurrence:

Exercise 9. Let $(X,{\mathcal F},T)$ be a topological dynamical system, and let x be a point in X.

1. Show that x is almost periodic if and and only if for every $p \in \beta {\Bbb Z}$ there exists $q \in \beta {\Bbb Z}$ such that $T^q T^p x = x$. (In particular, Lemma 1 is now an immediate consequence of Corollary 1.)
2. Show that x is recurrent if and only if there exists $p \in\beta {\Bbb Z} \backslash {\Bbb Z}$ such that $T^p x = x$. $\diamond$

Note that $\beta {\Bbb Z}$ acts on itself $\beta {\Bbb Z}$ by addition, $p: q \mapsto p+q$, with the action being continuous when p is an integer. Thus one can view $\beta {\Bbb Z}$ itself as a topological dynamical system, except with the caveat that $\beta {\Bbb Z}$ is not metrisable or even first countable (see Exercise 12). Nevertheless, it is still useful to think of $\beta {\Bbb Z}$ as behaving like a topological dynamical system. For instance:

Definition 1. An element $p \in \beta {\Bbb Z}$ is said to be minimal or almost periodic if for every $q \in \beta {\Bbb Z}$ there exists $r \in \beta {\Bbb Z}$ such that $r+q+p=p$.

Equivalently, p is minimal if $\beta {\Bbb Z}+p$ is a minimal left-ideal of $\beta {\Bbb Z}$, which explains the terminology.

Exercise 10. Show that for every $p \in \beta {\Bbb Z}$ there exists $q \in \beta {\Bbb Z}$ such that $q+p$ is minimal. (Hint: adapt the proof of Lemma 1 from the previous lecture.) Also, show that if p is minimal, then q+p and p+q are also minimal for any $q \in \beta {\Bbb Z}$. This shows that minimal elements of $\beta {\Bbb Z}$ exist in abundance. However, observe from Exercise 6 that no integer can be minimal. $\diamond$

Exercise 11. Show that if $p \in \beta {\Bbb Z}$ is minimal, and x is a point in a topological dynamical system $(X, {\mathcal F},T)$, then $T^p x$ is almost periodic. Conversely, show that x is almost periodic if and only if $x = T^p x$ for some minimal p. This gives an alternate (and more “algebraic”) proof of the Birkhoff recurrence theorem. $\diamond$

Exercise 12. Show that no element of $\beta {\Bbb Z} \backslash {\Bbb Z}$ can be written as a limit of a sequence in ${\Bbb Z}$. (Hint: if a sequence $n_j \in {\Bbb Z}$ converged to a limit $p \in \beta {\Bbb Z}$, one must have $\beta f(p) = \lim_{j \to \infty} f(n_j)$ for all functions $f: {\Bbb Z} \to K$ mapping into a compact Hausdorff space K.) Conclude in particular that $\beta {\Bbb Z}$ is not metrisable, first countable, or sequentially compact. $\diamond$

[Update, Jan 14: various corrections and reorganisation.]

[Update, Jan 30: Some exercises added or expanded.]

[Update, Feb 4: Hint for Exercise 12 added.]

[Update, Mar 21: Exercise 3 repaired.]