THE NUMBER OF TRANSLATES OF A CLOSED NOWHERE DENSE SET REQUIRED TO COVER A POLISH GROUP

ARNOLD W. MILLER AND JURIS STEPRĀNS

Abstract. For a Polish group 𝔾 let cov_{𝔾} be the minimal number of translates of a fixed closed nowhere dense subset of 𝔾 required to cover 𝔾. For many locally compact 𝔾 this cardinal is known to be consistently larger than cov(M) which is the smallest cardinality of a covering of the real line by meagre sets. It is shown that for several non-locally compact groups cov_{𝔾} = cov(M). For example the equality holds for the group of permutations of the integers, the additive group of a separable Banach space with an unconditional basis and the group of homeomorphisms of various compact spaces.

The notion of translation invariants corresponding to the usual invariants of the continuum has been considered by various researchers and an introductory survey can be found in §2.7 of the monograph [1] by Bartoszyński and Judah. The key definition for the purposes of this article is the cardinal they denoted by cov^{∗}(M). It is the least cardinal of a set X ⊆ ℝ such that there is some meagre set M ⊆ ℝ such that X + M = ℝ. It is asserted that the value of cov^{∗}(M) will be the same if the group (ℝ, +) is replaced in this definition by the Cantor set with its natural Boolean operation or an infinite product of finite cyclic groups. The goal of this note is to initiate a study of translation invariants for arbitrary Polish groups by establishing that not all Polish groups yield the same invariants and posing various questions which arise from this observation. Since many of the interesting questions in this area concern non-locally compact groups the measure-theoretic version is not easily formulated and, therefore, only the topological version will be considered.

Throughout, the statement that 𝔾 is a group will mean that 𝔾 = (G, ·,^{−1}) but for x and y in G the operation x · y will usually be abbreviated to xy. Similarly, if A ⊆ G and B ⊆ G then AB will denote the set {xy | x ∈ A and y ∈ B }. If A = {a} then {a}B will be abbreviated to aB.

To begin, a generalization of cov^{∗}(M) will be defined for arbitrary group actions.

Definition 1. Let 𝔾 be a group acting on a Polish space X with the action denoted by α : G × X → X. Define cov_{α} to be the least cardinal of a set Q ⊆ G such that there is some closed nowhere dense set C ⊆ X such that α(Q, C) = {α(h, c) | h ∈ Q and c ∈ C } = X. Define cov_{α}^{∗} to be the least cardinal of a set Q ⊆ G such that there is some meagre set M ⊆ X such that α(Q, M) = X. In the special case of a Polish group 𝔾 acting on itself by left translation these invariants will be denoted by cov_{𝔾} and cov_{𝔾}^{∗}. Note that the definitions of cov_{𝔾} and cov_{𝔾}^{∗} do not change if left translation is replaced by right translation since (QC)^{−1} = C^{−1}Q^{−1}.

1991 Mathematics Subject Classification. 03E17.

Key words and phrases. translations, nowhere dense sets, Polish groups, cardinal invariants.

Research for this paper was partially supported by NSERC of Canada.

The first observation applies to arbitrary groups.

Proposition 2. If 𝔾 is an arbitrary group, X is σ-compact, second countable and has no isolated points, α : G × X → X is a group action all of whose orbits are dense in X and α(g, ·) is continuous for each fixed g ∈ G, then cov_{α} = cov_{α}^{∗}.

Proof. It suffices to show that if M is meagre then there is a closed nowhere dense set C such that α(Y, C) ⊇ M for some countable set Y. Since X is σ-compact it may be assumed that M = ⋃_{n=0}^{∞} D_{n} where each D_{n} is compact and nowhere dense. Let {B_{n}}_{n=0}^{∞} be a base for X consisting of non-empty open sets. Select Y_{k} and U_{k} by induction on k such that

• Y_{k} is a finite subset of G

• U_{k} ⊆ B_{k} is a non-empty open set

• if j ≤ k and d ∈ D_{j} then there is some y ∈ Y_{j} such that α(y, d) ∉ ⋃_{i=0}^{k} U_{i} • the closure of ⋃_{j=0}^{k} U_{j} is not equal to X

If this can be done then let C be the complement of ⋃_{i=0}^{∞} U_{i} and take Y countable including ⋃_{n=0}^{∞} Y_{i} and closed under inverses. Then given x ∈ M there is some n such that x ∈ D_{n} and hence there exists y ∈ Y_{n} such that α(y, x) ∈ C and so

x = α(y^{−1}, α(y, x)) ∈ α({y^{−1}}, C) ⊆ α(Y, C).

To carry out the induction suppose this has been done for i < k. Since α is a continuous group action, α({g}, D_{j}) is closed nowhere dense for any g ∈ G and so it is possible to choose a non-empty open U_{k} ⊆ B_{k} disjoint from

⋃_{i<k} ⋃_{y∈Y_{i}} α({y}, D_{i}).

Moreover, it is possible to choose U_{k} so that there is some non-empty open set W witnessing the last induction clause; in other words, W ∩ ⋃_{i=0}^{k} U_{i} = ∅. Since D_{k} is compact and α({d}, G) is dense for each d ∈ D_{k} there is a finite set Y_{k} ⊆ G such that for each d ∈ D_{k} there is some y ∈ Y_{k} such that α(y, d) ∈ W. □

Proposition 3. cov_{𝔾} = cov_{𝔾}^{∗} for every non-discrete Polish 𝔾.

Proof. If 𝔾 is compact, then apply Proposition 2. On the other hand, Birkhoff and Kakutani proved that every Polish group admits a left invariant^{1} metric d inducing the same topology. While this metric d might not be complete, it is shown in Corollary 1.2.2 of [3] that the metric D defined by D(x, y) = d(x, y) + d(x^{−1}, y^{−1}) is a complete metric compatible with the topology. However D might not be left or right invariant. Nevertheless, since 𝔾 is not compact and D is complete it follows that D is not totally bounded; in other words, there is some ϵ > 0 and a sequence {x_{n}}_{n∈ω} such that D(x_{n}, x_{m}) > ϵ for all pairs of distinct integers n and m. A routine application of Ramsey’s Theorem yields an infinite subset S ⊆ ω such that either d(x_{n}, x_{m}) > ϵ/2 for all pairs of distinct integers n and m in S or d(x_{n}^{−1}, x_{m}^{−1}) > ϵ/2 for all pairs of distinct integers n and m in S. From the left invariance of the metric d it follows that letting U be the ϵ/3 ball around the identity either {x_{n}U}_{n∈ω} are pairwise disjoint sets or {x_{n}^{−1}U}_{n∈ω} are pairwise disjoint sets. Without loss of generality assume the former alternative.

1In other words, d(y, z) = d(xy, xz) for every x, y, z ∈ G. For a proof see [4] §7.1 page 115.

As in the argument for Proposition 2, it suffices to prove the following: For any family {C_{n}}_{n∈ω} of nowhere dense subsets of 𝔾 there exists a countable Q ⊆ 𝔾 and a nowhere dense C such that ⋃_{n∈ω} C_{n} ⊆ QC. Choose {q_{n}}_{n∈ω} ⊆ 𝔾 so that ⋃_{n∈ω} q_{n}U = 𝔾. This is possible since {xU | x ∈ G} covers G and G is Lindelof. Let k : ω × ω → ω be a bijection and define r_{n,m} so that r_{n,m}q_{n} = x_{k(n,m)} and define

C = ⋃_{m∈ω} ⋃_{n∈ω} r_{n,m}(q_{n}U ∩ C_{m}).

Note that r_{n,m}(q_{n}U ∩ C_{m}) ⊆ x_{k(n,m)}U and since {x_{n}U}_{n∈ω} are disjoint, C is nowhere dense. On the other hand if Q = {r_{n,m}^{−1}}_{n,m∈ω}, then QC contains q_{n}U ∩ C_{m} for each n and m and hence C_{m} ⊆ QC. □

The proof of the following fact can be found in [1] in Lemma 2.4.2.

Theorem 4.

cov(M) = min {|F| | F ⊆ ω^{ω} and ∀g ∈ ω^{ω} ∃f ∈ F ∀n f(n) ≠ g(n)} .

Corollary 5. cov_{ℤ^{ω}} = cov(M).

Proof. For any non-discrete Polish group 𝔾 the inequality cov_{𝔾} ≥ cov(M) holds. Let

C = {f ∈ ℤ^{ω} | ∀n f(n) ≠ 0} .

Note that C is closed nowhere dense. Take F ⊆ ω^{ω} ⊆ ℤ^{ω} such that |F| = cov(M) and such that

∀g ∈ ℤ^{ω} ∃f ∈ F ∀n f(n) ≠ g(n).

But this means that F + C = ℤ^{ω}. Hence cov_{ℤ^{ω}} ≤ |F| = cov(M). □

To generalize this to other groups the following lemma will play a key role. It is stated using standard notation about trees. If T ⊆ ^{⌣^{ω}}ω is a tree then

[T] = {f | ∀n ∈ ω f ↾ n ∈ T }

and T is infinite branching if for each t ∈ T the set Σ(t) = {n ∈ ω | t^{⌢}(n) ∈ T } is infinite.

Lemma 6. If T ⊆ ^{⌣^{ω}}ω is an infinite branching tree then there is F such that • |F| = cov(M)

• F ⊆ [T]

• for each g : ω → ω there is f ∈ F such that f(n) ≠ g(n) for all integers n.

Proof. Using Theorem 4 let F^{∗} ⊆ ω^{ω} be a family of cardinality cov(M) such that for every g ∈ ω^{ω} there is f ∈ F^{∗} such that f(n) ≠ g(n) for every integer n. By choosing an infinite branching subtree of T if necessary, it may be assumed that Σ(t) ∩ Σ(s) = ∅ unless s = t. Let ψ_{t} : ω → Σ(t) be a bijection for each t ∈ T. For any f : ω → ω define f_{ψ} by inductively setting

f_{ψ}(n) = ψ_{f_{ψ}↾n}(f(n))

and note that f_{ψ} ↾ n ∈ T for all n. Then let F = {f_{ψ} | f ∈ F^{∗} }.

Let Σ = ∪_{s∈T} Σ(s). Given g : ω → ω and k ∈ ω if g(k) ∈ Σ let s_{k} ∈ T be the unique node in T with g(k) ∈ Σ(s_{k}).

Now define G : ω → ω by

G(k) = \begin{cases}ψ_{s_{k}}^{−1}(g(k)) & if g(k) ∈ Σ \\ 0 & otherwise \end{cases}

and choose f ∈ F^{∗} such that f(n) ≠ G(n) for all n. It suffices to show that f_{ψ}(n) ≠ g(n) for all n. Assume that f_{ψ}(n) = g(n). Then necessarily g(n) = f_{ψ}(n) ∈ Σ(f_{ψ} ↾ n) and so s_{n} = f_{ψ} ↾ n and therefore

G(n) = ψ_{s_{n}}^{−1}(g(n)) = ψ_{f_{ψ}↾n}^{−1}(f_{ψ}(n)) = f(n)

which is a contradiction.

□

Theorem 7. Let 𝔾 be a Polish group such that there are {B_{k}, A_{k}^{j}}_{j,k∈ω} such that: (1) B_{k} and A_{k}^{j} are all subsets of G

(2) there is an infinite branching tree T ⊆ ^{⌣^{ω}}ω such that ⋂_{k∈ω} A_{k}^{b(k)} ≠ ∅ for b ∈ [T]

(3) ⋃_{k∈ω} B_{k} is dense open

(4) (A_{k}^{i}B_{k}) ∩ (A_{k}^{j}B_{k}) = ∅ unless i = j then cov_{𝔾} = cov(M).

Proof. Using Lemma 6 let F be a set of branches through T such that |F| = cov(M) and for each g : ω → ω there is f ∈ F such that f(n) ≠ g(n) for all integers n. For b ∈ [T] let b^{∗} ∈ ⋂_{k} A_{k}^{b(k)} and let X = {b^{∗} | b ∈ F }. Let C = G \ ⋃_{k} B_{k}. Then C is closed and nowhere dense and |X| = cov(M). Hence it suffices to show that XC = G.

To this end let g ∈ G and define Γ : ω → ω such that g ∈ A_{n}^{Γ(n)}B_{n} if there is any i with g ∈ A_{n}^{i}B_{n}. By Hypothesis 3 the choice of Γ(n) is unique if it exists at all. Let f ∈ F be such that f(n) ≠ Γ(n) for all integers n. If g ∉ XC then g ∉ f^{∗}C and so there is some integer m such that g ∈ f^{∗}B_{m} ⊆ A_{m}^{f(m)}B_{m}. Hence Γ(m) = f(m) which is impossible. □

For example, for the group ℤ^{ω} one could take A_{k}^{j} = {α | α(k) = j }, B_{k} = A_{k}^{0}, and T = ^{⌣^{ω}}ω or, for the group ℝ^{ω} take B_{k} = {α ∈ ℝ^{ω} | |α(k)| < 1/3} and let A_{k}^{j} and T be the same.

Corollary 8. If 𝔾 is the group of all permutations of the integers then cov_{𝔾} = cov(M).

Proof. Let A_{i}^{k} = {p | p(2i) = 2k } and B_{k} = A_{k}^{k}. Letting T be the tree of all one-to-one sequences satisfies the hypotheses of Theorem 7. □

For any compact metric space X let ℍ(X) be the group of autohomeomorphisms of X using composition as the group operation and the topology induced by the uniform metric.

Corollary 9. cov_{ℍ([0,1])} = cov(M).

Proof. Construct a family of open intervals {I_{s} | s ∈ ℤ^{<ω} } so that:

• I_{∅} = (0, 1)

• I_{s} = ⋃_{n∈ℤ} \overline{I_{s^{⌢}(n)}} and the right-hand endpoint of I_{s^{⌢}(n)} is the left-hand end-

point of I_{s^{⌢}(n+1)}
• the length of I_{s} is less than \frac{1}{|s|}.

In other words, I_{s} is partitioned into contiguous subintervals in order type ℤ and their endpoints. Define for i, k ∈ ω the open sets

U_{k}^{i} = ∪ {I_{s} | s ∈ ℤ^{k+1} and s(k) = i}

Fix any p_{0} ∈ (0, 1). Define

B_{k} = {h ∈ ℍ | h(p_{0}) ∈ U_{k}^{0} }

and let U = ⋃_{k∈ω} U_{k}^{0} and note that U is open dense and

⋃_{k∈ω} B_{k} = {h ∈ ℍ | h(p_{0}) ∈ U }

is open dense in ℍ. For any k, i ∈ ω define

A_{k}^{i} = {h ∈ ℍ | h(U_{k}^{0}) ⊆ U_{k}^{i} } .

It is clear that {A_{k}^{i} ∘ B_{k}}_{i∈ω} are pairwise disjoint because if h ∈ A_{k}^{i} ∘ B_{k} then h(p_{0}) ∈ U_{k}^{i} and {U_{k}^{i}}_{i∈ω} are pairwise disjoint.

Now let T = ^{⌣^{ω}}ω and suppose b ∈ [T]. In order to find h ∈ ⋂_{k∈ω} A_{k}^{b(k)} let a_{s} denote the left-hand endpoint of I_{s} and define Q = {a_{s} | s ∈ ℤ^{<ω} }. Note that Q is dense in [0, 1] and its ordering is exactly the same as the lexicographical ordering <_{lex} on ℤ^{<ω}: Define s <_{lex} t if and only if s ⊂ t or there exists i such that s(i) < t(i) and s ↾ i = t ↾ i. Then a_{s} < a_{t} if and only if s <_{lex} t. for all s, t ∈ ℤ^{<ω}.

For each s ∈ ℤ^{n} define s + b = t ∈ ℤ^{n} by pointwise addition and define h(a_{s}) = a_{s+b}. Clearly the mapping s ↦ s + b is a bijection preserving the lexicographical order and so h : Q → Q is an order preserving bijection. Thus it extends uniquely to an order preserving bijection h^{∗} on [0, 1]. Note that h(I_{s^{⌢}(0)}) ⊆ I_{t^{⌢}(b(k))} for any t, s and b where t = s + b and k = |s|. So h^{∗}(U_{k}^{0}) ⊆ U_{k}^{b(k)} for each k. Thus h^{∗} ∈ ⋂_{k∈ω} A_{k}^{b(k)}. □

Corollary 10. Suppose there exists π : X → [0, 1] which is (1) continuous, onto, and open

(2) ∀h ∈ ℍ([0, 1]) ∃ĥ ∈ ℍ(X) ∀x ∈ X π(ĥ(x)) = h(π(x))

(3) if U ⊆ (0, 1) is open dense, then there exists p_{0} ∈ X such that

{h ∈ ℍ(X) | h(p_{0}) ∈ π^{−1}(U)}

is open dense then cov_{ℍ(X)} = cov(M).

Proof. Using the same set U = ⋃_{k∈ω} U_{k}^{0} ⊆ (0, 1) as defined in the proof of Corollary 9 let p_{0} ∈ X be such that

{h ∈ ℍ(X) | h(p_{0}) ∈ π^{−1}(U)}

is open dense. Define

B_{k} = {h ∈ ℍ(X) | π ∘ h(p_{0}) ∈ U_{k}^{0} }

and note that

⋃_{k∈ω} B_{k} = {h ∈ ℍ(X) | h(p_{0}) ∈ π^{−1}(U)}

is open dense in ℍ(X). For any integers k and i define

A_{k}^{i}(X) = {h ∈ ℍ(X) | π ∘ h (π^{−1}(U_{k}^{0})) ⊆ U_{k}^{i} }

reserving the notation A_{k}^{i} for the sets defined in the proof of Corollary 9. It is clear that {A_{k}^{i}(X) ∘ B_{k}}_{i∈ω} are pairwise disjoint because if h ∈ A_{k}^{i}(X) ∘ B_{k} then π ∘ h(p_{0}) ∈ U_{k}^{i} and {U_{k}^{i}}_{i∈ω} are pairwise disjoint.

Now let T = ^{⌣^{ω}}ω and suppose b ∈ [T]. It has already been established in the proof of Corollary 9 that there is h ∈ ⋂_{k∈ω} A_{k}^{b(k)}. Then it is readily verified that ĥ ∈ ⋂_{k∈ω} A_{k}^{b(k)}(X). □

Corollary 11. If X is either [0, 1]^{k} for k ∈ ω or the infinite dimensional Hilbert cube [0, 1]^{ω} then cov_{ℍ(X)} = cov(M).

Proof. Use Corollary 10 applied to the mapping which projects a sequence to its first coordinate. Given h ∈ ℍ([0, 1]) take ĥ ∈ ℍ([0, 1]^{k}) to be defined by

ĥ(x_{0}, x_{1}, x_{2}, …) = (h(x_{0}), x_{1}, x_{2}, …).

Let p_{0} = (1/2, 1/2, …, 1/2) and note that by a classical result of L.E.J. Brouwer (invariance of domain) h(p_{0}) is in the interior of [0, 1]^{k} and so its first coordinate is neither 0 or 1. It is easy to see that condition (3) of Corollary 10 holds. In the case of [0, 1]^{ω} it is possible that the first coordinate of h(p_{0}) is 0 or 1 since the Hilbert cube is homogeneous. However, one of the standard proofs of homogeneity, see Lemma 6.1.4 page 252 in Van Mill [8], shows that given any (u_{n} : n ∈ ω) ∈ [0, 1]^{ω}, there exists a small k ∈ hom([0, 1]^{ω}) with k(u_{n} : n ∈ ω) = (v_{n} : n ∈ ω) and v_{0} ∈ (0, 1). □

Since the homeomorphism group of the Hilbert cube is universal — in the sense that any Polish group is homeomorphic to one of its closed subgroups — the following question is of central importance in this area.

Question 12. If 𝔾 is a non-σ-compact, closed subgroup of ℍ([0, 1]^{ω}) does the equality cov_{𝔾} = cov(M) hold?

Corollary 13. If S^{n} is the n dimensional Euclidean sphere then

cov_{ℍ(S^{n})} = cov(M).

Similarly for any compact metric space X we have that

cov_{ℍ(S^{n}×X)} = cov(M).

Proof. Use Corollary 10 applied to the mapping which projects the sphere onto one of its diameters. □

Corollary 14. If 𝔾 is the additive group of a separable Banach space with an unconditional basis then cov_{𝔾} = cov(M).

Proof. Let {e_{i}}_{i∈ω} be an unconditional basis for the Banach space 𝔹. Recall that this implies that for each S ⊆ ω the projection map

P_{S} (∑_{n<ω} α_{n}e_{n}) = ∑_{n∈S} α_{n}e_{n}

is a well-defined continuous linear operator.

Let {S_{j}}_{j∈ω} be pairwise disjoint infinite subsets of ω. Choose {δ_{i}}_{i∈ω} positive reals such that ∑_{i=0}^{∞} δ_{i} < ∞. For each k the range of P_{S_{k}} is an infinite dimensional Banach space and hence the ball of diameter δ_{k} around the origin is not compact and therefore not totally bounded. In other words, it is possible to find some ϵ_{k} > 0 and {u_{k}^{i}}_{i∈ω} contained in the range of P_{S_{k}} such that ∥u_{k}^{i}∥ < δ_{k} and ∥u_{k}^{i} − u_{k}^{j}∥ > ϵ_{k} for distinct i and j. Let

B_{k} = {b ∈ 𝔹 | ∥P_{S_{k}}(b)∥ < ϵ_{k}/2} and A_{k}^{j} = {b ∈ 𝔹 | P_{S_{k}}(b) = u_{k}^{j} } .

Finally, let T = ⋃_{n<ω} ∏_{k<n} S_{k} and note that T is infinite branching.

To check that the other hypotheses of Theorem 7 are satisfied note that B_{k} is open because P_{S_{k}} is continuous. To see that ⋃_{k} B_{k} is dense note that the finite linear sums of the basis form a dense subset and if b = ∑_{i=0}^{N} α_{i}e_{i} and S_{k} is disjoint from {0, 1, …, N} then P_{S_{k}}(b) = 0. To see that ⋂_{k} A_{k}^{b(k)} is non-empty for b ∈ [T] note that b^{∗} = ∑_{k=0}^{∞} u_{k}^{b(k)} is a convergent series by the choice of the δ_{i}. The sets {A_{k}^{j} + B_{k}}_{j∈ω} are pairwise disjoint by the choice of ϵ_{k} □

Question 15. Is Corollary 14 true for Banach spaces without an unconditional basic sequence?

Definition 16. For f : ℕ → ℕ define 𝔢𝔮(f) to be the least cardinal of a set X ⊆ ∏_{i=0}^{∞} f(i) such that for all g ∈ ∏_{i=0}^{∞} f(i) there is x ∈ X such that x(i) ≠ g(i) for every i ∈ ℕ. Let 𝔢𝔮 be the minimum of all 𝔢𝔮(f) such that lim_{i→∞} f(i) = ∞.

Proposition 17. (1) cov_{𝔾} = 𝔢𝔮 when 𝔾 is either the product of finite cyclic groups or ℝ.

(2) Let 𝔾 = 𝕋 = ℝ/ℤ be the circle group or any finite dimensional torus 𝕋^{n}. Or let 𝔾 product of countably many nontrivial finite groups. Then cov_{𝔾} = 𝔢𝔮.

Proof. Assertion (1) is proved in [1] and their argument generalizes to (2). □

Lemma 18. Suppose 𝔾 and ℍ are topological groups and h : 𝔾 → ℍ is a homomorphism onto H.

(a) If h is open and continuous, then cov_{ℍ} ≥ cov_{𝔾}.

(b) If h takes meagre sets to meagre sets, then cov_{ℍ}^{∗} ≤ cov_{𝔾}^{∗}.

Proof. (a) Since h is open and continuous, the pre-image of a nowhere dense set is nowhere dense. Let C ⊆ ℍ be nowhere dense and Y ⊆ ℍ have the property that |Y | = cov_{ℍ} and Y C = H. Then B = h^{−1}(C) is nowhere dense in 𝔾 and choose Y so that h(Y) = X and |X| = |Y |. Since h is homomorphism it follows that h(Y B) = XC = H. In order to see that Y B = G note that for any g ∈ G there exists y ∈ Y and b ∈ B with h(g) = h(yb). It follows that h(y^{−1}g) = h(b) ∈ C so y^{−1}g ∈ B and so g = yb′ for some b′ ∈ B.

(b) Suppose h takes meagre sets to meagre sets and let M ⊆ 𝔾 be meagre such that cov_{𝔾}^{∗} = |X| for some X with XM = G. Then h(X)h(M) = H and so cov_{ℍ}^{∗} ≤ |X| = cov_{𝔾}^{∗}

□

So, for example, if 𝔾 and ℍ are Polish groups then cov_{ℍ×𝔾} ≤ cov_{𝔾}.

Proposition 19. For any integer n the equality cov_{ℝ^{n}} = 𝔢𝔮 holds.

Proof. The quotient mapping from ℝ^{n} to ℝ^{n}/ℤ^{n} is both open and it takes nowhere dense sets to meagre sets. It follows from Lemma 18 that cov_{ℝ^{n}} ≤ cov_{ℝ^{n}/ℤ^{n}} and cov_{ℝ^{n}}^{∗} ≥ cov_{ℝ^{n}/ℤ^{n}}^{∗}.

Since ℝ^{n}/ℤ^{n} is isomorphic to 𝕋^{n} it follows from Proposition 17 and Corollary 3 that cov_{ℝ^{n}} = 𝔢𝔮. □

There are many models of set theory where cov(M) < 𝔢𝔮; examples can be found using modified Silver forcing as in §7 of [6].

Theorem 20. It is relatively consistent with ZFC that • 2^{ℵ_{0}} = ℵ_{2},

• cov_{𝔾} = ℵ_{2} for every infinite compact group 𝔾

• and cov(M) = ℵ_{1}. In fact, the last equality will from the fact that 𝔡 = ℵ_{1} holds in the model constructed.

Proof. Fix 𝔾 an infinite compact group and C ⊆ G closed nowhere dense. Let h : 2^{ω} → G be a continuous onto map. Recall the notation preceding Lemma 6 concerning trees. If p is a tree and s ∈ p define p_{s} = {t ∈ p | t ⊆ s or s ⊆ t}.

Definition 21. Define the partially ordered set ℙ = ℙ_{(h,C,𝔾)} as follows: p ∈ ℙ if and only if p ⊆ ^{⌣^{ω}}2 is a perfect tree such that xC is relatively nowhere dense in h([p_{s}]) for every x ∈ G and s ∈ p. This is equivalent to saying for every x ∈ G and s ∈ p there exists t ∈ p with s ⊆ t and h([p_{t}]) ∩ xC = ∅. The ordering on ℙ is inclusion considering a subtree as a stronger condition.

First note that the forcing is not vacuous. For s ∈ ^{⌣^{ω}}2 let B_{s} = {x ∈ 2^{ω} | s ⊆ x}. If p = {s ∈ ^{⌣^{ω}}2 | h(B_{s}) is non-meagre in G} then p ∈ ℙ. It will be shown that the forcing has the following three properties:

(a) ℙ_{(h,C,𝔾)} is a proper forcing notion.

(b) Forcing with ℙ_{(h,C,𝔾)} adds no ≤^{∗} unbounded sequence in ω^{ω}.

(c) In the generic extension there exists z ∈ G such that z ∉ XC where X is the set of elements of G in the ground model.

Claim 1. Given any p ∈ ℙ and n < ω there exists N with n < N < ω such that for every s ∈ 2^{n} ∩ p and x ∈ G there exists t ∈ 2^{N} ∩ p_{s} such that h([p_{t}]) ∩ xC = ∅.

Proof. To prove this claim, fix s ∈ p and x ∈ G. Since xC is nowhere dense in h[p_{s}] there exists s_{x} ∈ p with s ⊆ s_{x} and h[p_{s_{x}}] ∩ xC = ∅. By the continuity of the group operation and the compactness of xC, there exists an open neighbourhood U_{x} of x with h[p_{s_{x}}]∩U_{x}C = ∅. By compactness, finitely many of these U_{x} cover and so there is some N larger than the length of the finitely many corresponding s_{x} for each of the finitely many s in 2^{n} ∩ p. □

Now suppose a sequence (p^{n}, k_{n}) has been constructed satisfying for every n < ω:

(1) p^{n} ∈ ℙ,

(2) k_{n} < k_{n+1} < ω,

(3) p^{n+1} ≤ p^{n},

(4) 2^{k_{n}} ∩ p^{n+1} = 2^{k_{n}} ∩ p^{n},

(5) any s ∈ 2^{k_{n}} ∩ p^{n+1} has incomparable extensions in 2^{k_{n+1}} ∩ p^{n+1}, and

(6) for each s ∈ p^{n+1} ∩ 2^{k_{n}} and x ∈ G there exists t ∈ p^{n+1} ∩ 2^{k_{n+1}} with s ⊆ t

and h([p_{t}^{n+1}]) ∩ xC = ∅. Then p = ⋂_{n∈ω} p^{n} is in ℙ. Except for the last condition 6 the fusion conditions described are identical to those used by Baumgartner and Laver in [2] for Sacks forcing. The last condition guarantees that for every x ∈ G and s ∈ p that xC is nowhere dense in h([p_{s}]).

In order to verify the properties (a), (b), and (c) note that this fusion property is close enough to a property-A forcing to see easily that ℙ is proper. Next it will be shown that forcing with ℙ is ω^{ω}-bounding. Suppose

p^{0} ⊩ “τ : ω → Ordinals”.

Construct a fusion sequence p^{n}, k_{n} as follows. At stage n given p^{n}, k_{n} find p^{n+1} ≤ p^{n} such that for every s ∈ 2^{k_{n}} ∩ p^{n} the condition p^{n+1} decides τ(n). Hence there exists a finite set F_{n} such that

p^{n+1} ⊩ “τ(n) ∈ F̌_{n}”.

Now pick k_{n+1} large enough to satisfy Condition 5 and Condition 6. The fusion p = ⋂_{n∈ω} p^{n} then satisfies:

p ⊩ “∀n τ(n) ∈ F̌_{n}”.

Lastly, if G is ℙ_{(h,C,𝔾)}-generic over a ground model V and x ∈ 2^{ω} is the real determined by G — in other words, x ∈ [p] for all p ∈ G — then z = h(x) has the property that z ∉ XC where X = V ∩ G. This follows from any easy density argument, since if p ∈ ℙ and s ∈ p then p_{s} ∈ ℙ.

To prove the Theorem note that the countable support iteration of proper forcings which are ω^{ω}-bounding is ω^{ω}-bounding (Shelah, see [1] 6.3.5). Hence a countable support ω_{2}-iteration over a model of GCH will satisfy 𝔡 = ℵ_{1}. By dovetailing it is easy to arrange that every infinite compact group and closed nowhere dense subset in the final model is forced with (or rather a code for such) at unboundedly many stages in the iteration. It follows by standard arguments that in the final model cov_{𝔾} = ℵ_{2} for every infinite compact group 𝔾.

□

Question 22. Is it consistent to have a compact group 𝔾 such that cov_{𝔾} > 𝔢𝔮?

Question 23. Is it true that for any infinite compact group 𝔾 that cov_{𝔾} ≥ 𝔢𝔮?

Question 24. Is it true that for every non-discrete Polish group 𝔾 that cov_{𝔾} = 𝔢𝔮 or cov_{𝔾} = cov(M)?

The question of general group actions has only been hinted at but there are many questions concerning these as well. For example, it is easy to see that if α_{n} is the natural action of the isometry group on ℝ^{n} and n ≤ m then cov_{α_{n}} ≤ cov_{α_{m}}. However the following question is unanswered.

Question 25. Is it true that cov_{α_{n}} = cov_{α_{m}} for all m and n?

References

[1] Tomek Bartoszyński and Haim Judah. Set theory. On the structure of the real line. A K Peters Ltd., Wellesley, MA, 1995.

[2] James E. Baumgartner and Richard Laver. Iterated perfect-set forcing. Ann. Math. Logic, 17(3):271–288, 1979.

[3] Howard Becker and Alexander S. Kechris. The descriptive set theory of Polish group actions, volume 232 of London Mathematical Society Lecture Note Series. Cambridge University Press, Cambridge, 1996.

[4] Greg Hjorth. Classification and orbit equivalence relations, volume 75 of Mathematical Surveys and Monographs. American Mathematical Society, Providence, RI, 2000.

[5] Lindenstrauss, Joram; Tzafriri, Lior. Classical Banach spaces. I. Sequence spaces. Ergebnisse der Mathematik und ihrer Grenzgebiete, Vol. 92. Springer-Verlag, Berlin-New York, 1977. xiii+188 pp. ISBN: 3-540-08072-4

[6] Arnold W. Miller. Some properties of measure and category. Trans. Amer. Math. Soc., 266(1):93–114, 1981.

[7] Arnold W. Miller. Corrections and additions to: “Some properties of measure and category”. Trans. Amer. Math. Soc., 271(1):347–348, 1982.

[8] van Mill, J.; Infinite-dimensional topology. Prerequisites and introduction. North-Holland Mathematical Library, 43. North-Holland Publishing Co., Amsterdam, 1989. xii+401 pp. ISBN: 0-444-87133-0.

Appendix

This is not intended for publication but only for the electronic version.

Details of the proof of Proposition 17 (2).

Proposition 26. If 𝔾_{n} is a nontrivial finite group for each n ∈ ℕ and 𝔾 = ∏_{n=0}^{∞} 𝔾_{n} with the product topology then cov_{𝔾} = 𝔢𝔮.

Proof. To begin, note that if f ≤ g then 𝔢𝔮(f) ≥ 𝔢𝔮(g). Choose f : ℕ → ℕ such that 𝔢𝔮(f) = 𝔢𝔮. Now let {k_{n}}_{n=0}^{∞} be an increasing sequence of integers such that g(n) = ∏_{i=k_{n}}^{k_{n+1}−1} |G_{i}| ≥ f(n) for each n. Clearly 𝔢𝔮(g) = 𝔢𝔮 as well. Now let C = {x ∈ G | (∀n ∈ ℕ)(∃i ∈ [k_{n}, k_{n+1})) x(i) ≠ e_{i} } and note that C is closed nowhere dense. Using the definition of 𝔢𝔮(g) it is possible to find X ⊆ G such that for all y ∈ G there is x ∈ X such that y ↾ [k_{n}, k_{n+1}) ≠ x ↾ [k_{n}, k_{n+1}) for all n. Therefore ∏_{n=0}^{∞}(x(n)y(n)) ∈ C; in other words XC = G and so cov_{𝔾} ≤ 𝔢𝔮.

On the other hand, suppose that C ⊆ G is closed nowhere dense and there is W ⊆ G such that |W | < 𝔢𝔮 and W C = G. Then it is possible to find disjoint intervals of integers {I_{n}}_{n=0}^{∞} such that, letting π_{n} : G → ∏_{i∈I_{n}} G_{i} be the projection map, for all n there is x_{n} ∈ ∏_{i∈I_{n}} G_{i} such that π_{n}^{−1}{x_{n}} ∩ C = ∅. Let

C^{∗} = {y ∈ ∏_{n=0}^{∞} ∏_{i∈I_{n}} G_{i} | (∀n) y(n) ≠ x_{n} }

and let g(n) = ∏_{i∈I_{n}} |G_{i}|. Let {g_{m}}_{m=1}^{g(n)} enumerate ∏_{i∈I_{n}} G_{i}. For w ∈ W let f_{w} : ℕ → ℕ be defined by f_{w}(n) = j if and only if x_{n}π_{n}(w) = g_{j}.

Since |W | < 𝔢𝔮(g) it is possible to find h : ℕ → ℕ such that for all w ∈ W there is some n ∈ ℕ such that f_{w}(n) = h(n). Let z = ∏_{n=0}^{∞} g_{h(n)}. It suffices to note that zπ_{n}(w)^{−1} ∉ C^{∗} for all w ∈ W because it then follows that π^{−1}{z}{w^{−1}} ∩ C = ∅ for all w ∈ W and so W C ≠ G. To see that zπ_{n}(w)^{−1} ∉ C^{∗} let n be such that h(n) = f_{w}(n). Then x_{n}π_{n}(w) = g_{h(n)} = z ↾ I_{n}. Hence zπ_{n}(w)^{−1} = x_{n} and so zπ_{n}(w)^{−1} ∉ C^{∗}. □

Proposition 27. For any integer n the equality cov_{𝕋^{n}} = 𝔢𝔮 holds.

Proof. While there is a continuous and open mapping from 2^{ℕ} onto the circle 𝕋, this mapping is not a homomorphism. Nevertheless, it is sufficiently close to a homomorphism to be able to do the proof.

This follows the pattern of Proposition 26 taking care of carry digits. To be more precise, given ψ : ℕ → ℕ such that 𝔢𝔮(ψ) = 𝔢𝔮 let {k_{n}}_{n=0}^{∞} be an increasing sequence of integers such that ∏_{i=k_{n}}^{k_{n+1}−1} |G_{i}| ≥ 2ψ(n) for each n. Consider 𝕋 to be ℝ/ℤ so that 𝕋 is identified with [0, 1]. For t ∈ [0, 1] let x_{t} : ℕ → 2 be chosen so that ∑_{i=0}^{∞} x_{t}(i)/2^{i} = t and there are infinitely many i such that x_{t}(i) = 0. Now let

C = {t ∈ [0, 1] | (∀n ∈ ℕ)(∃i ∈ [k_{n}, k_{n+1})) x(i) ≠ 0}

and note that C is closed nowhere dense. Now, for any two functions f : [k_{n}, k_{n+1} − 1) → 2 and g : [k_{n}, k_{n+1} − 1) → 2 define f ≡_{n} g if and only if there is some j such that

• f(i) ≠ g(i) for i ≥ j
• f(i) = g(i) for i < j
• f ↾ [j + 1, k_{n+1}) is constant ( and hence so is g ↾ [j + 1, k_{n+1})).

It is immediate that ≡_{n} is an equivalence relation whose equivalence classes are all pairs. Let S_{n} be the set of equivalence classes of ≡_{n}. Hence ϕ(n) = |S_{n}| ≥ ψ(n).

Using the definition of 𝔢𝔮(ϕ) it is possible to find Y ⊆ ∏_{n=0}^{∞} S_{n} such that for all t ∈ [0, 1] there is y ∈ Y such that x_{t} ↾ [k_{n}, k_{n+1}) ≢_{n} y(n) for all n. For each y ∈ Y let y^{∗} be a choice function which selects an element of y(n) for each n and let ȳ = ∑_{n=0}^{∞} ∑_{i=k_{n}} k_{n+1} − 1y^{∗}(i)/2^{i}. Observe that x_{t} ↾ [k_{n}, k_{n+1}) ≢_{n} xȳ ↾ [k_{n}, k_{n+1}) and hence, regardless of whether digits are carried from the right, x_{t−ȳ} ↾ [k_{n}, k_{n+1}) is not identically 0. In other words, t ∈ ȳ + C.

On the other hand, suppose that C ⊆ [0, 1] is closed nowhere dense and there is W ⊆ [0, 1] such that |W | < 𝔢𝔮 and W +C = [0, 1]. Then it is possible to find disjoint intervals of integers {I_{n}}_{n=0}^{∞} such that, letting ≡_{n} be the equivalence relation defined on functions from I_{n} to 2 and S_{n} be as before, for all n there is [s_{n}]_{≡_{n}} ∈ S_{n} such that

{t ∈ [0, 1] | x_{t} ↾ I_{n} ≡_{n} s_{n} } ∩ C = ∅.

Let

C^{∗} = {t ∈ [0, 1] | (∀n) x_{t} ↾ I_{n} ≢_{n} s_{n} }

and let g(n) = 2^{|I_{n}|}. Let {g_{m}}_{m=1}^{g(n)} enumerate 2^{I_{n}}. For w ∈ W let f_{w} : ℕ → ℕ be defined by f_{w}(n) = j if and only if g_{j} − s_{n} = x_{w} ↾ I_{n}.

Since |W | < 𝔢𝔮(g) it is possible to find h : ℕ → ℕ such that for all w ∈ W there is some n ∈ ℕ such that f_{w}(n) = h(n). Now argue as in Proposition 26. □

The following is another example of a corollary to 10.

Corollary 28. If X is an orientable 2-manifold then cov_{ℍ(X)} = cov(M).

Proof. An orientable 2-manifold is homeomorphic to the surface of the unit cube with holes drilled along axes perpendicular to the xy-plane. Use Corollary 10 applied to the mapping which projects the surface to the z-axis. □

Department of Mathematics, University of Wisconsin, Van Vleck Hall, 480 Lincoln Drive, Madison, Wisconsin, USA 53706-1388

  E-mail address: miller@math.wisc.edu                    www.math.wisc.edu/∼miller

Department of Mathematics, York University, 4700 Keele Street, Toronto, Ontario, Canada M3J 1P3

  E-mail address: steprans@yorku.ca                     www.math.yorku.ca/∼steprans