Ideals with Smital properties

Abstract

A $\sigma$-ideal $\mathcal {I}$ on a Polish group $(X,+)$ has the Smital Property if for every dense set D and a Borel $\mathcal {I}$-positive set B the algebraic sum $D+B$ is a complement of a set from $\mathcal {I}$. We consider several variants of this property and study their connections with the countable chain condition, maximality and how well they are preserved via Fubini products. In particular we show that there are $\mathfrak {c}$ many maximal invariant $\sigma$-ideals with Borel bases on the Cantor space $2^\omega$.

1 Introduction

We adopt the usual set-theoretical notation. We say that X is a Polish space if it is a separable and completely metrisable topological space. $\text {Bor}(X)$ denotes the family of Borel subsets of X. $\mathcal {M}(X)$ and $\mathcal {N}(X)$ are the families of meager and null subsets of X respectively. Sometimes we will write briefly $\mathcal {M}$ and $\mathcal {N}$ if the underlying space is clear from the context.

Let $\mathcal {A}$ be a ($\sigma$-)algebra and $\mathcal {I}$ be a ($\sigma$-)ideal on an Abelian Polish group $(X,+)$. Throughout the paper we assume that $\mathcal {I}$ contains all singletons and

$$\begin{aligned} (\forall I\in \mathcal {I}) (\exists A\in \mathcal {A}\cap \mathcal {I}) (I\subseteq A). \end{aligned}$$

If $\mathcal {A}$ is not explicitly stated we assume $\mathcal {A}=\text {Bor}(X)$. In such a case $\mathcal {I}$ has a Borel base.

We say that a set A is $\mathcal {I}$-positive, if $A\notin \mathcal {I}$. A is called $\mathcal {I}$-residual if $A^c\in \mathcal {I}$, we denote this fact by $A\in \mathcal {I}^\star$.

For any sets $A, B\subseteq X$ we denote the algebraic sum of these sets by $A+B$, i.e.

$$\begin{aligned} A+B=\{a+b: a\in A,\; b\in B\}. \end{aligned}$$

Let us now recall the classical notion of the Steinhaus property and probably less famous notions of Smital properties which were studied in [2].

Definition 1

We say that a pair $(\mathcal {A},\mathcal {I})$ has

• (i) the Steinhaus Property if for any $A,B\in \mathcal {A}\backslash \mathcal {I}$ the set $A-B$ has a nonempty interior;

• (ii) the Smital Property, briefly SP, if for every dense set D and every $A\in \mathcal {A}\backslash \mathcal {I}$ the set $A+D$ is $\mathcal {I}$-residual;

• (iii) the Weaker Smital Property, briefly WSP, if there exists a countable and dense set D such that for every $A\in \mathcal {A}\backslash \mathcal {I}$ the set $A+D$ is $\mathcal {I}$-residual;

• (iv) the Very Weak Smital Property, briefly VWSP, if for every $A\in \mathcal {A}\backslash \mathcal {I}$ there is a countable set D such that the set $A+D$ is $\mathcal {I}$-residual.

Note that $\mathcal {M}$ and $\mathcal {N}$ have all of these properties.

The following Proposition seems to be folklore but we could not find the proof of the second part in the literature.

Proposition 2

The Steinhaus Property is equivalent to SP.

Proof

The Steinahus Property implies SP. Let $A\in \mathcal {A}\backslash \mathcal {I}$ and D be countable and dense. We may assume that D is a subgroup. Suppose that $(A+D)\notin \mathcal {I}^\star$. Then $(A+D)^c\notin \mathcal {I}$. By the Steinhaus Property $(A+D)-(A+D)^c$ contains an open neighborhood of 0. A contradiction since $0\notin (A+D)-(A+D)^c$.

SP implies the Steinhaus Property. Assume that $A-B$ has an empty interior for $A,B\in \mathcal {A}\backslash \mathcal {I}$. Then there is a countable dense set $D\subseteq (A-B)^c$. It follows from SP that $(B+D)\cap A\ne \emptyset$, a contradiction. $\square$

Let $F\subseteq X\times Y$. Then for $x\in X$

$$\begin{aligned} F_x=\{y\in Y:\; (x,y)\in F\} \end{aligned}$$

is the vertical section of F at x. Similarly, for $y\in Y$

$$\begin{aligned} F^y=\{x\in X:\; (x,y)\in F \} \end{aligned}$$

is the horizontal section of F at y.

For ($\sigma$-)algebras $\mathcal {A}\subseteq P(X)$ and $\mathcal {B}\subseteq P(Y)$ let $\mathcal {A}\otimes \mathcal {B}\subseteq P(X\times Y)$ denote the ($\sigma$-)algebra generated by the rectangles of the form $A\times B$ with $A\in \mathcal {A}$ and $B\in \mathcal {B}$.

Definition 3

Let $(\mathcal {A},\mathcal {I})$ and $(\mathcal {B},\mathcal {J})$ be pairs of ($\sigma$-)algebra - ($\sigma$-)ideal on Polish spaces X and Y respectively. Then we define the Fubini product of $\mathcal {I}$ and $\mathcal {J}$ as follows:

$$\begin{aligned} K\in \mathcal {I}\otimes \mathcal {J}\Leftrightarrow (\exists C\in \mathcal {A}\otimes \mathcal {B})(K\subseteq C \wedge \{x\in X: C_x\notin \mathcal {J}\}\in \mathcal {I}). \end{aligned}$$

Notice that in the case of ideals possessing Borel bases, i.e. $\mathcal {A}=\text {Bor}(X)$ and $\mathcal {B}=\text {Bor}(Y)$ the above definition ensures the existence of a Borel base for $\mathcal {I}\otimes \mathcal {J}$.

Proposition 4

If $(\mathcal {A}\otimes \mathcal {B},\mathcal {I}\otimes \mathcal {J})$ has SP (WSP, VWSP), then $(\mathcal {A},\mathcal {I})$ and $(\mathcal {B},\mathcal {J})$ also have it.

Proof

Let us consider the case of WSP for $\mathcal {I}$. Let D be a witness that $\mathcal {I}\otimes \mathcal {J}$ has WSP. Let $A\in \mathcal {A}\backslash \mathcal {I}$. The set $R=D+(A\times Y)$ is $\mathcal {I}\otimes \mathcal {J}$-residual, therefore

$$\begin{aligned} \widetilde{R}=\{x\in X: R_x \text { is }\mathcal {J}-\text {residual}\} \end{aligned}$$

is $\mathcal {I}$-residual. Clearly, $A+\pi _X(D)=\widetilde{R}$, hence we are done. $\square$

The following definition is a variation of the ones found in [7, Definition 18.5], [9] and agrees with the notation given in [3].

Definition 5

Let X and Y be Polish spaces and let $\mathcal {F}\subseteq P(X)$, $\mathcal {G}\subseteq P(Y)$, $\mathcal {H}\subseteq P(X\times Y)$ be families of sets. Then we say that $\mathcal {G}$ is $\mathcal {H}$-on-$\mathcal {F}$ if for each set $H\in \mathcal {H}$

$$\begin{aligned} \{x\in X: H_x\in \mathcal {G}\}\in \mathcal {F}. \end{aligned}$$

Mainly we will be interested in the case where ${\mathcal {G}=\mathcal {J}\subseteq P(Y)}$ is a $\sigma$-ideal, $\mathcal {F}\in \{Bor(X), \sigma (Bor(X)\cup \mathcal {I})\}$ and $\mathcal {H}\in \{Bor(X\times Y), \sigma (Bor(X\times Y)\cup \mathcal {I}\otimes \mathcal {J})\}$. Here $\sigma (\mathcal {D})$ is the $\sigma$-algebra generated by the family $\mathcal {D}$. We will write, for example, that $\mathcal {J}$ is Borel-on-measurable instead of $Bor(X\times Y)$-on-$\sigma (Bor(X)\cup \mathcal {I})$ if the context is clear. Notice that both $\mathcal {M}$ and $\mathcal {N}$ are Borel-on-Borel (see [7, Exercise 22.22, 22.25]).

Example 6

Measurable-on-measurable not necessarily implies Borel-on-Borel.

Proof

Let $\mathcal {J}=\{\emptyset \}$ and take a Borel set B projection of which is analytic and not Borel. $\square$

Proposition 7

Borel-on-measurable implies measurable-on-measurable.

Proof

Assume that $\mathcal {J}$ is Borel on measurable. Let $C\subseteq X\times Y$ be measurable with respect to $\mathcal {I}\otimes \mathcal {J}$. Then $C=(B\backslash A_1)\cup A_2$, where B is Borel and $A_1, A_2\in \mathcal {I}\otimes \mathcal {J}$. Clearly

$$\begin{aligned} \{x\in X: {A_2}_x\notin \mathcal {J}\}\in \mathcal {I}, \end{aligned}$$

hence it is measurable. See that

$$\begin{aligned}&\{x\in X: (B\backslash A_1)_x\notin \mathcal {J}\}=\{x\in X: B_x\backslash {A_1}_x\notin \mathcal {J}\}\\&\quad =\{x\in X: B_x\notin \mathcal {J}, {A_1}_x\in \mathcal {J}\}\cup \{x\in X: B_x\notin \mathcal {J}, {A_1}_x\notin \mathcal {J}, B_x\backslash {A_1}_x\notin \mathcal {J}\}. \end{aligned}$$

Since $\{x\in X: B_x\notin \mathcal {J}\}$ is measurable, $\{x\in X: {A_1}_x\in \mathcal {J}\}\in \mathcal {I}^\star$ and

$$\begin{aligned} \left\{ x\in X: B_x\notin \mathcal {J}, {A_1}_x\notin \mathcal {J}, B_x\backslash {A_1}_x\notin \mathcal {J}\right\} \subseteq \{x\in X: {A_1}_x\notin \mathcal {J}\}\in \mathcal {I}, \end{aligned}$$

the set $\{x\in X: C_x\notin \mathcal {J}\}$ is measurable. $\square$

2 Smital and ccc

Let $\mathcal {I}$ be a $\sigma$-ideal in a Polish space X and assume that $\mathcal {I}$ has a Borel base. We say that $\mathcal {I}$ satisfies the countable chain condition (briefly: ccc) if every family of pairwise disjoint Borel $\mathcal {I}$-positive sets is countable. Let us also recall the following cardinal coefficient

$$\begin{aligned} \text {cov}(\mathcal {I})=\min \left\{ |\mathcal {A}|: \mathcal {A}\subseteq \mathcal {I},\, \bigcup \mathcal {A}=X\right\} . \end{aligned}$$

Theorem 8

Let $\mathcal {I}$ be a $\sigma$-ideal possessing WSP. Then $\mathcal {I}$ satisfies $\text {ccc}$ or $\text {cov}(\mathcal {I})=\omega _1$.

Proof

Let $\mathcal {I}$ be $\sigma$-ideal with WSP and let D witness it. Let $\{B_\alpha : \alpha <\omega _1\}$ be a family of pairwise disjoint Borel $\mathcal {I}$-positive sets. WSP implies that for each $\alpha <\omega _1$ a set $D+B_\alpha$ is $\mathcal {I}$-residual. If $\bigcap _{\alpha <\omega _1}(D+B_\alpha )=\emptyset$, then $\text {cov}(\mathcal {I})=\omega _1$. On the other hand, if $\bigcap _{\alpha <\omega _1}(D+B_\alpha )\ne \emptyset$ then for $x\in \bigcap _{\alpha <\omega _1}(D+B_\alpha )$ we have

$$\begin{aligned} (\forall \alpha \in \omega _1) (\exists d\in D)( x\in d+B_\alpha ). \end{aligned}$$

D is countable, hence there exist a set $W\subseteq \omega _1$ of cardinality $\omega _1$ and $d\in D$ such that

$$\begin{aligned} (\forall \alpha \in W)( x\in d+B_\alpha ), \end{aligned}$$

which gives $x-d\in \bigcap _{\alpha \in W}B_\alpha$, a contradiction. $\square$

The following remark improves the result obtained in [4].

Remark 9

Let $\mathcal {I}$ be a $\sigma$-ideal possessing WSP. Then the following statements are equivalent:

• (i) For each family of sets $\{B_\alpha : \alpha <\omega _1\}\subseteq \mathcal {B}\backslash \mathcal {I}$ there exists a set $W\subseteq \omega _1$ of cardinality $\omega _1$ such that $\bigcap _{\alpha \in W}B_\alpha \ne \emptyset$;

• (ii) $\text {cov}(\mathcal {I})>\omega _1$.

Proof

(ii) $\Rightarrow$ (i) is a part of Theorem 8. To prove (i) $\Rightarrow$ (ii) let us suppose that $\text {cov}(\mathcal {I})=\omega _1$. Then there is a family of sets

$$\begin{aligned} \{A_\alpha : \alpha <\omega _1\}\subseteq \mathcal {B}\cap \mathcal {I}\end{aligned}$$

for which $\bigcup _{\alpha <\omega _1}A_\alpha =X$. Set $\widetilde{A}_\alpha =\bigcup _{\beta \le \alpha }A_\alpha$ for each $\alpha <\omega _1$. The family $\{\widetilde{A}_\alpha : \alpha <\omega _1\}$ is ascending and covers X. Hence $\{\widetilde{A}_{\alpha }^c: \alpha <\omega _1\}$ is descending family of $\mathcal {I}$-residual sets. Moreover, for every $W\subseteq \omega _1$ of cardinality $\omega _1$ we have $\bigcap _{\alpha \in W}\widetilde{A}_{\alpha }^c=\emptyset$, which contradicts (i). $\square$

3 Preserving Smital properties via products

In [2] the authors present some results on various Smital properites in product spaces. Their setup is, in their words, as general as possible, concerned with algebras and ideals. It is not clear if they intended their results to hold for $\sigma$-algebras and $\sigma$-ideals or algebras and ideals only. The formulation of [2, Theorem 4.2] suggests the former since it is concerned with the Borel algebra and the families of meager and null sets. In their proof they rely implicitly on the following property.

Definition 10

Let $\mathcal {A}\!\subseteq \! P(X\!\times \! Y)$ be a ($\sigma$-)algebra and let $\mathcal {I}\!\subseteq \! P(X\!\times \! Y)$ be a ($\sigma$-)ideal. A pair $(\mathcal {A}, \mathcal {I})$ has the positive rectangle property (PRP) if for every $\mathcal {I}$-positive set $A\in \mathcal {A}$ there is an $\mathcal {I}$-positive rectangle R satisfying $R\subseteq A\cup I$ for some $I\in \mathcal {I}$.

In this section we show explicitly that PRP holds for pairs algebra-ideal. However, PRP does not hold for pairs $\sigma$-algebra - $\sigma$-ideal in general, including the relevant here pair of Borel $\sigma$-algebra and the family of null sets.

Example 11

The pair $(\text {Bor}(\mathbb {R}^2), [\mathbb {R}]^{\le \omega }\otimes [\mathbb {R}]^{\le \omega })$ does not have PRP.

Proof

Let $P\subseteq \mathbb {R}$ be a perfect set such that $P\cap (P+x)$ is at most 1-point for $x\ne 0$ (see [8]). Let us set

$$\begin{aligned} B=\{(x,y): x\in P \wedge y\in P-x\}\backslash (\mathbb {R}\times \{0\}). \end{aligned}$$

B is Borel and $B\notin [\mathbb {R}]^{\le \omega }\otimes [\mathbb {R}]^{\le \omega }$. If $x\in B^y$, then $x\in P$ and $x\in P-y$, therefore $B^y$ is at most 1-point.

Let us suppose that there are sets $A_1, A_2\in \text {Bor}(\mathbb {R}^2)\backslash [\mathbb {R}]^{\le \omega }$ and a set $K\in [\mathbb {R}]^{\le \omega }\otimes [\mathbb {R}]^{\le \omega }$ such that $(A_1\times A_2)\backslash K\subseteq B$. Let $T=\{x\in \mathbb {R}: |K_x|>\omega \}$ and notice that $|T|\le \omega$. Pick $x_0\in A_1\backslash T$. Then $K_{x_{0}}$ is countable. By the definition of B for each $y\in A_2\backslash K_{x_0}$ we have $K^y\supseteq A_1\backslash (\{x_0\}\cup T)$. So $A_2\backslash K_{x_0}\subseteq K_{x_1}$ for $x_1\in A_1\backslash (\{x_0\}\cup T)$, a contradiction. $\square$

It is clear that the pair $(\text {Bor}(\mathbb {R}^2), \mathcal {M})$ has PRP, since every nonmeager set possessing the property of Baire is nonempty and open, modulo a set of the first category. What about $(\text {Bor}(\mathbb {R}^2), \mathcal {N})$? As a warm up let us recall the following folklore result.

Proposition 12

Every set $E\subseteq [0,1]^2$ of positive measure contains a subset of the same measure which does not contain a rectangle of positive measure.

Proof

Let $E\subseteq [0,1]^2$ have positive measure. Consider $E'=\{(x,y)\in E: x-y\in \mathbb {Q}^c\}$. To see that $\lambda (E')=\lambda (E)$ let us observe that $E'_x=E_x\cap (\mathbb {Q}^c+x)$ for every $x\in [0,1]$ and $\mathbb {Q}^c$ is co-null. Now, if $A\times B\subseteq E'$, A and B of positive measure, then $A-B$ should contain a nonempty open set (Steinhaus Theorem), but clearly $\mathbb {Q}\cap (A-B)=\emptyset$. A contradiction. This completes the proof. $\square$

This result may be improved with the following Lemma.

Lemma 13

There exists a set $F\subseteq \mathbb {R}$ such that $\lambda (F\cap U)>0$ and $\lambda (F^c\cap U)>0$ for every nonempty open set U.

Proof

Let $(B_n: n\in \omega )$ be an enumeration of the basis of $\mathbb {R}$. At the step 0 let $C^1_0, C^2_0 \subseteq B_0$ be two disjoint Cantor sets of positive measure. At the step $n+1$ let assume that we have two sequences of pairwise disjoint Cantor sets $(C^1_k: k\le n)$ and $(C^2_k: k\le n)$ which for all $i\in \{0,1\}$ and $k\le n$ satisfy $\lambda (C^i_k\cap B_k)>0$. The set

$$\begin{aligned} B_{n+1}\backslash \bigcup _{k\le n}(C^1_k\cup C^2_k) \end{aligned}$$

is nonempty and open, hence it contains two disjoint Cantor sets of positive measure. Denote them by $C^1_{n+1}$ and $C^2_{n+1}$. This completes the construction and $F=\bigcup _{n\in \omega }C^1_k$ is the desired set. $\square$

The above Lemma will serve as a tool to prove the result from [6].

Example 14

(Erdös, Oxtoby). There is a set $E\subseteq \mathbb {R}^2$ such that $E\cap (A\times B)$ and $E^c\cap (A\times B)$ have positive measure for each $A, B\subseteq \mathbb {R}$ of positive measure.

Proof

Let F be as in the formulation of Lemma 13 and set ${E=\{(x,y)\in \mathbb {R}^2: x-y\in F\}}$. Let $A, B\subseteq \mathbb {R}$ have a positive measure. Then

$$\begin{aligned} \lambda (E\cap A\times B)&=\iint \chi _{F}(x-y)\chi _{A}(x)\chi _{B}(y)dxdy\\&= \iint \chi _{F}(x)\chi _A(x+y)\chi _{B}(y)dxdy\\&=\int _F \lambda ((A-x)\cap B)dx. \end{aligned}$$

$\lambda ((A-x)\cap B)$ is a continuous non-negative function. Furthermore, it is positive on some interval, since $\int _\mathbb {R}\lambda ((A-x)\cap B)dx=\lambda (A\times B)$, so $\int _F \lambda ((A-x)\cap B)dx>0$. $\square$

Corollary 15

$(\text {Bor}({\mathbb {R}^2}), \mathcal {N})$ does not have PRP.

Now we will focus on PRP for products of algebras.

Lemma 16

Let $\mathcal {A}\subseteq P(X)$ and $\mathcal {B}\subseteq P(Y)$ be algebras. Then $\mathcal {A}\otimes \mathcal {B}\subseteq P(X\times Y)$ consists of finite unions of rectangles.

Proof

First let us observe that complements of rectangles are finite unions of rectangles:

$$\begin{aligned} (A\times B)^c=(A\times B^c)\cup (A^c\times B)\cup (A^c\times B^c). \end{aligned}$$

Next, see that finite intersection of a finite union of rectangles is again a finite union of rectangles. Let $T_1$, $T_2\subseteq \omega$ be finite. Let $A^n_k$, $B^n_k$ be rectangles from $\mathcal {A}$ and $\mathcal {B}$ respectively for every $n\in T_1$ and $k\in T_2$. Then

$$\begin{aligned} \bigcap _{n\in T_1}\bigcup _{k\in T_2}A^n_k\times B^n_k=\bigcup _{f\in T_2^{T_1}}\bigcap _{n\in T_1}A^{n}_{f(n)}\times B^n_{f(n)}. \end{aligned}$$

$T_2^{T_1}$ is finite and finite intersections of rectangles are also rectangles, hence the proof is complete. $\square$

For algebra $\mathcal {A}$ and ideal $\mathcal {I}$ let $\mathcal {A}[\mathcal {I}]$ denote the algebra generated by $\mathcal {A}\cup \mathcal {I}$.

Proposition 17

Let $\mathcal {A}\subseteq P(X)$ and $\mathcal {B}\subseteq P(Y)$ be algebras and let $\mathcal {I}$ be an ideal in $X\times Y$. Then $((\mathcal {A}\otimes \mathcal {B})[\mathcal {I}],\mathcal {I})$ has PRP.

Proof

Notice that $(\mathcal {A}\otimes \mathcal {B})[\mathcal {I}]=\{C\triangle I: C\in \mathcal {A}\otimes \mathcal {B},\; I\in \mathcal {I}\}$. PRP follows from the previous Lemma. $\square$

From now on let $\mathcal {A}\subseteq P(X), \mathcal {B}\subseteq P(Y)$ be $\sigma$-algebras, and $\mathcal {I}\subseteq P(X)$, $\mathcal {J}\subseteq P(Y)$ $\sigma$-ideals.

Theorem 18

Let $\mathcal {I}$ and $\mathcal {J}$ possess WSP and assume one of the following properties

• (i) $\mathcal {J}$ is Borel-on-Borel;

• (ii) $\mathcal {J}$ measurable-on-measurable;

• (iii) $(\text {Bor}(X\times Y), \mathcal {I}\otimes \mathcal {J})$ has PRP.

Then $\mathcal {I}\otimes \mathcal {J}$ also has WSP.

Proof

Let $D_1$ and $D_2$ witness WSP for $\mathcal {I}$ and $\mathcal {J}$ respectively. Let $B\in \text {Bor}(X\times Y)\backslash \mathcal {I}\otimes \mathcal {J}$. If any of the properties (i)-(iii) holds then a set $\widetilde{B}=\{x\in X: B_x\notin \mathcal {J}\}$ contains a Borel, $\mathcal {I}$-positive set and $D_1+\widetilde{B}$ is $\mathcal {I}$-residual. Let us observe that

$$\begin{aligned} (D_1\times D_2)+B\supseteq \bigcup _{d_1\in D_1}\bigcup _{x\in \widetilde{B}}(\{d_1+x\}\times (D_2+B_x)), \end{aligned}$$

therefore for every $x\in D_1+\widetilde{B}$ the set $((D_1\times D_2)+B)_x$ is $\mathcal {J}$-residual. Since $D_1+\widetilde{B}$ is $\mathcal {I}$-residual, the proof is complete. $\square$

In [1] the authors showed that $\mathcal {M}\otimes \mathcal {N}$ and $\mathcal {N}\otimes \mathcal {M}$ have SP and thus WSP. The following corollary extends this result regarding WSP.

Corollary 19

Let $n\in \omega$ and $\mathcal {I}_k\in \{\mathcal {M}, \mathcal {N}\}$ for any $k\le n$. Then $\mathcal {I}_0\otimes \mathcal {I}_1 \otimes ... \otimes \mathcal {I}_n$ has WSP.

Proof

By [5, Lemma 3.1] the ideal $\mathcal {I}_0\otimes \mathcal {I}_1 \otimes ... \otimes \mathcal {I}_n$ is Borel-on-Borel for any $n\in \omega$ and $\mathcal {I}_k\in \{\mathcal {M}, \mathcal {N}\}$, $k\le n$. $\square$

In [2, Theorem 4.3] the authors showed that if $\mathcal {B}=\mathcal {J}\cup \mathcal {J}^\star$ and $(\mathcal {A}, \mathcal {I})$ has SP then $(A\otimes \mathcal {B}, \mathcal {I}\otimes \mathcal {J})$ also has SP. We will generalize this result (Theorem 24). Let us start with two technical definitions.

Definition 20

We say that a pair $(\mathcal {A}\otimes \mathcal {B}, \mathcal {I}\otimes \mathcal {J})$ has the Tall Rectangle Hull Property (TRHP) if for every set $C\in \mathcal {A}\otimes \mathcal {B}$

$$\begin{aligned} (\exists \widetilde{C}\in \mathcal {A}, I \in \mathcal {I}, J \in \mathcal {J})((\widetilde{C}\backslash I)\times (Y\backslash J)\subseteq C \subseteq (\widetilde{C}\times Y)\cup (I \times Y)\cup (X \times J)). \end{aligned}$$

If a set C fulfills the above condition we will say that it has TRHP witnessed by the triple $(\widetilde{C}, I, J)$.

Analogously we define Wide Rectangle Hull Property (WRHP):

$$\begin{aligned} (\exists \widetilde{C}\in \mathcal {B}, I \in \mathcal {I}, J \in \mathcal {J})((X\backslash I)\times (\widetilde{C}\backslash J)\subseteq C \subseteq (X\times \widetilde{C})\cup (I \times Y)\cup (X \times J)). \end{aligned}$$

Proposition 21

If a pair $(\mathcal {A}\otimes \mathcal {B}, \mathcal {I}\otimes \mathcal {J})$ have TRHP or WRHP then it has PRP.

Proof

Let $C\in \mathcal {A}\otimes \mathcal {B}$ has TRHP witnessed by $(\widetilde{C}, I, J)$:

$$\begin{aligned} (\widetilde{C}\backslash I)\times (Y\backslash J)\subseteq C \subseteq (\widetilde{C}\times Y)\cup (I \times Y)\cup (X \times J)) \end{aligned}$$

and assume that $C\notin \mathcal {I}\otimes \mathcal {J}$. Then $\widetilde{C}\times Y$ is the desired rectangle. Clearly $\widetilde{C}\times Y\subseteq C$ modulo a set from $\mathcal {I}\otimes \mathcal {J}$. It is also $\mathcal {I}\otimes \mathcal {J}$-positive, otherwise

$$\begin{aligned} (\widetilde{C}\times Y)\cup (I \times Y)\cup (X \times J))\in \mathcal {I}\otimes \mathcal {J}\end{aligned}$$

and also $C\in \mathcal {I}\otimes \mathcal {J}$.

The proof of WRHP case is almost identical. $\square$

Lemma 22

The family of sets possessing TRHP is closed under countable unions and complements. The same is true for the family of sets possessing WRHP.

Proof

Proofs for both cases follow the same pattern, so without loss of generality let us focus on the case of TRHP.

Let $C=\bigcup _{n\in \omega }C_n$ and $(\widetilde{C}_n, I_n, J_n)$ witness TRHP for $C_n$, $n\in \omega$. Then for each $n \in \omega$

$$\begin{aligned} (\widetilde{C}_n\backslash I_n)\times (Y\backslash J_n)\subseteq C_n \subseteq (\widetilde{C}_n\times Y)\cup (I_n\times Y)\cup ( X\times J_n). \end{aligned}$$

Then

$$\begin{aligned} \left( \bigcup _{n\in \omega }\widetilde{C}_n\backslash \bigcup _{n\in \omega }I_n\right) \times \left( Y\backslash \bigcup _{n\in \omega }J_n\right)&\subseteq \bigcup _{n\in \omega }\left( \widetilde{C}_n\backslash I_n\right) \times (Y\backslash J_n)\subseteq \bigcup _{n\in \omega }C_n=C \\&\subseteq \bigcup _{n\in \omega }\bigg ((\widetilde{C}_n\times Y)\cup (I_n\times Y)\cup ( X\times J_n)\bigg ) \\&\subseteq \left( \left( \bigcup _{n\in \omega }C_n\right) \times Y\right) \cup \left( \left( \bigcup _{n\in \omega }I_n\right) \times Y\right) \\ {}&\cup \left( X\times \left( \bigcup _{n\in \omega }J_n\right) \right) . \end{aligned}$$

Hence, setting $\widetilde{C}=\bigcup _{n\in \omega }\widetilde{C}_n$, $I=\bigcup _{n\in \omega }I_n$, $J=\bigcup _{n\in \omega }J_n$ completes this part of the proof.

Now let $C=D^c$ for D witnessing TRHP with $(\widetilde{D}, I, J)$. We have

$$\begin{aligned} (\widetilde{D}\backslash I) \times (Y\backslash J)\subseteq D \subseteq (\widetilde{D}\times Y) \cup (I\times Y) \cup (X\times J). \end{aligned}$$

Through complementation

$$\begin{aligned} \big ((\widetilde{D}\times Y) \cup (I\times Y) \cup (X\times J)\big )^c\subseteq C \subseteq \big ((\widetilde{D}\backslash I) \times (Y\backslash J)\big )^c. \end{aligned}$$

Let us focus on the right-hand side

$$\begin{aligned} \big ((\widetilde{D}\backslash I) \times (Y\backslash J)\big )^c= & {} ((\widetilde{D}\backslash I)\times (Y\backslash J)^c) \cup ((\widetilde{D}\backslash I)^c\times Y)\\\subseteq & {} ((\widetilde{D})^c\times Y)\cup (X\times J)\cup (I\times Y). \end{aligned}$$

Now the left-hand side. It is an intersection of the following sets

$$\begin{aligned} (\widetilde{D}\times Y)^c&=(\widetilde{D})^c\times Y, \\ (I\times Y)^c&=(X\backslash I) \times Y, \\ (X\times J)^c&=X\times (Y\backslash J), \end{aligned}$$

which is equal to

$$\begin{aligned} ((\widetilde{D})^c\backslash I) \times (Y\backslash J). \end{aligned}$$

In summary

$$\begin{aligned} ((\widetilde{D})^c\backslash I) \times (Y\backslash J)\subseteq C \subseteq ((\widetilde{D})^c\times Y)\cup (X\times J)\cup (I\times Y). \end{aligned}$$

Then $((\widetilde{D})^c, I, J)$ witnesses TRHP for C. The proof is complete. $\square$

Theorem 23

Let $\mathcal {C}=\mathcal {A}\otimes \mathcal {B}$. Then

• (i) if $\mathcal {A}=\mathcal {I}\cup \mathcal {I}^\star$ then $(\mathcal {C}, \mathcal {I}\otimes \mathcal {J})$ has WRHP.

• (ii) if $\mathcal {B}=\mathcal {J}\cup \mathcal {J}^\star$ then $(\mathcal {C}, \mathcal {I}\otimes \mathcal {J})$ has TRHP.

Proof

Let us prove (ii) (the proof of (i) is similar). Let us notice that rectangles from $\mathcal {C}$ have TRHP. Indeed, if $C=A\times B$ then set

$$\begin{aligned} \widetilde{C}&=A,\; I=\emptyset ,\; J=B^c \text { if } B\in \mathcal {J}^\star \\ \widetilde{C}&=\emptyset ,\; I=\emptyset ,\; J=B \text { if } B\in \mathcal {J}. \end{aligned}$$

The rest of the proof relies on Lemma 22 which allows us to perform an induction over the hierarchy of sets making up the $\sigma$-algebra $\mathcal {C}$. $\square$

Theorem 24

Let $\mathcal {C}=\mathcal {A}\otimes \mathcal {B}$ and assume that

• (i) $(\mathcal {C}, \mathcal {I}\otimes \mathcal {J})$ has TRHP and $(\mathcal {A}, \mathcal {I})$ has SP, or

• (ii) $(\mathcal {C}, \mathcal {I}\otimes \mathcal {J})$ has WRHP and $(\mathcal {B}, \mathcal {J})$ has SP.

Then $(\mathcal {C},\mathcal {I}\otimes \mathcal {J})$ has SP.

Proof

Assume (i). Let $D\subseteq X\times Y$ be dense, set $D_1=\pi _1(D)$ and let $B\in \mathcal {C}$ be $\mathcal {I}\otimes \mathcal {J}$-positive. Then there are $\widetilde{B}\in \mathcal {A}\backslash \mathcal {I}$ and $J\in \mathcal {J}$ such that $\widetilde{B}\times (Y\backslash J)\subseteq B$. It follows that

$$\begin{aligned} D+B\supseteq D+(\widetilde{B}\times (Y\backslash J))\supseteq \bigcup _{d_1\in D_1}\bigcup _{d_2\in D_{d_1}}(d_1+\widetilde{B})\times (d_2+Y\backslash J)). \end{aligned}$$

Therefore for every $x\in D_1+\widetilde{B}$ the set $(D+B)_x$ contains a translation of $Y\backslash J$. By SP $D_1+\widetilde{B}\in \mathcal {I}^{\star }$ thus $D+B$ is $\mathcal {I}\otimes \mathcal {J}$-residual.

The proof assuming (ii) is analogous. $\square$

4 Maximal invariant $\sigma$-ideals with Borel bases

There is a surprising connection between maximal invariant $\sigma$-ideals with Borel bases and Smital properties.

Proposition 25

The following are equivalent:

• (i) $\mathcal {I}$ has VWSP;

• (ii) $\mathcal {I}$ is maximal among invariant proper $\sigma$-ideals with Borel base.

Proof

• $(i)\Rightarrow (ii):$ Let us suppose that $\mathcal {J}\supsetneq \mathcal {I}$ is such an ideal. Let A be a Borel set from $\mathcal {J}\backslash \mathcal {I}$. Then there exists a countable set D such that $D+A$ is $\mathcal {I}$-residual, therefore $\mathcal {J}$-residual, hence $\mathcal {J}$ is not proper.

• $(ii)\Rightarrow (i):$ Let us suppose that there is a set $B\in Bor\backslash \mathcal {I}$ for which $B+D$ is not $\mathcal {I}$-residual for every countable set D. Consider the family

  $$\begin{aligned} \mathcal {I}'=\{A\cup C: A\in \mathcal {I}\wedge (\exists D)(C\subseteq D+B \wedge |D|\le \omega )\}. \end{aligned}$$

It is an invariant proper $\sigma$-ideal satisfying $\mathcal {I}\subsetneq \mathcal {I}'$, which leads to a contradiction. $\square$

Proposition 26

Let $\{\mathcal {I}_n: n\in \omega \}$ be a countable family of pairwise distinct maximal invariant $\sigma$-ideals on X with Borel bases. Then for each $n\in \omega$ the $\sigma$-ideal $\mathcal {I}_n$ is orthogonal to $\bigcap _{k\in \omega \backslash \{n\}}\mathcal {I}_k$.

Proof

Fix $n\in \omega$. There are sets $A_k\in (\text {Bor}(X)\cap \mathcal {I}_n)\backslash \mathcal {I}_k$, $k\in \omega \backslash \{n\}$. By Proposition 25 for each of them there is a countable set $C_k$ such that $A_k+C_k\in \mathcal {I}_n\cap \mathcal {I}_k^\star$. Therefore

$$\begin{aligned} \bigcup _{k\in \omega \backslash \{n\}}(A_k+C_k)\in \mathcal {I}_n\cap \bigcap _{k\in \omega \backslash \{n\}}\mathcal {I}_k^\star . \end{aligned}$$

$\square$

Corollary 27

Let $\mathcal {I}$ be a maximal invariant $\sigma$-ideal with a Borel base different from $\mathcal {M}$ and $\mathcal {N}$. Then there is $A\in \mathcal {I}\cap (\mathcal {M}\cap \mathcal {N})^\star$.

Let us now focus on $X=2^\omega$. The following result incorporates techniques similar to these used in [10, Theorem 3.1].

Theorem 28

There are $\mathfrak {c}$ many maximal invariant $\sigma$-ideals on $2^\omega$.

Proof

Let $\{A_\alpha : \alpha <\mathfrak {c}\}$ be an AD family on $\omega$, i.e. for all distinct $\alpha , \beta <\mathfrak {c}$ the set $A_\alpha \cap A_\beta$ is finite. For every $\alpha <\mathfrak {c}$ set

$$\begin{aligned} \mathcal {I}_\alpha =\{A\subseteq 2^\omega :\;&(\exists B\in \text {Bor}(2^\omega )A\subseteq B\wedge (\exists M\in \mathcal {M}(2^{A_\alpha }))(\forall x\in 2^{A_\alpha }) \\&(x\notin M \rightarrow B^{\alpha }_x\in \mathcal {N}(2^{\omega \backslash A_\alpha }))\}, \end{aligned}$$

where $B^\alpha _x=\{y\restriction \omega \backslash A_\alpha :\; y\in B,\; y\restriction A_\alpha =x\}$.

Let us show that $I_\alpha \ne I_\beta$ if $\alpha \ne \beta$. Set

$$\begin{aligned}&M\in \mathcal {M}(2^{A_\alpha \backslash A_\beta })\backslash \mathcal {N}(2^{A_\alpha \backslash A_\beta }), \\&C=\{x\in 2^\omega : x\restriction (A_\alpha \backslash A_\beta )\in M\}. \end{aligned}$$

Notice that $C\in I_\alpha$. The set

$$\begin{aligned} M_\alpha =\bigcup _{t\in 2^{A_\alpha \cap A_\beta }}\left\{ y\in 2^{A_\alpha }: y\restriction (A_{\alpha }\cap A_{\beta })=t\wedge y\restriction (A_\alpha \backslash A_\beta )\in M\right\} \end{aligned}$$

is a finite union of meager sets, hence meager. For each $x\in 2^{A_\alpha }\backslash M_\alpha$ the set $C^\alpha _x$ is empty.

On the other hand for each $x\in 2^{A_\beta }$

$$\begin{aligned} C^\beta _x=\left\{ y\in 2^{\omega \backslash A_\beta }: y\restriction (A_\alpha \backslash A_\beta )\in M\right\} . \end{aligned}$$

The above set may be considered as a product $M\times 2^{\omega \backslash (A_\alpha \cup A_\beta )}$ of non-null set and the whole space, which is not null.

Now we will show that every $\mathcal {I}_\alpha$ is maximal among invariant $\sigma$-ideals on $2^\omega$ with Borel bases. Each $\mathcal {I}_\alpha$ is essentially $\mathcal {M}(2^{A_\alpha })\otimes \mathcal {N}(2^{\omega \backslash A_\alpha })$. It follows that $\mathcal {I}_\alpha$ has WSP, thus by Proposition 25 the proof is complete. $\square$

The reasoning in the above Theorem does not translate to the case of $\mathbb {R}$. We may ask the following question.

Question 29

Are $\mathcal {M}$ and $\mathcal {N}$ the only maximal invariant $\sigma$-ideals with Borel bases in $\mathbb {R}$?

Question 30

Is it true that for every set $G\in (\mathcal {M}\cap \mathcal {N})^\star$ there is a countable set C such that $C+G=\mathbb {R}$?

A positive answer to Question 30 would also answer positively Question 29.