Borel’s conjecture and meager-additive sets

It is clear that if $\mathbf{P}$ satisfies Axiom $\mathsf{B}$, then it satisfies Baumgartner’s Axiom $\mathsf{A}$ (see [4, §7] for definitions), and therefore is proper.

To see that $\mathbf{P}$ is $\omega^{\omega}$-bounding, let $G$ be a $V$-generic filter on $\mathbf{P}$, let $f\colon\mathbf{N}\to\mathbf{N}$ be a function in $V[G]$, and set $p_{0}:=\textbf{1}_{\mathbf{P}}$. For each $n\in\mathbf{N}$, let $H_{n}\in V$ be a finite set, and let $p_{n+1}\leq_{n}p_{n}$ be such that $p_{n+1}\Vdash``f(n)\in H_{n}"$. Let $g\colon\mathbf{N}\to\mathbf{N}$ be the function in $V$ defined by $g(n):=\max H_{n}+1$. If $p\leq_{n}p_{n}$ for every $n\in\mathbf{N}$, then

and therefore, in $V[G]$, $f(n)<g(n)$ for all $n\in\mathbf{N}$. ∎

Follows from Theorem III.3.2 and Theorem V.4.3 in [21]. ∎

Every forcing notion satisfying Axiom $\mathsf{B}$ is strongly $\omega^{\omega}$-bounding in the sense of Goldstern, Judah, and Shelah in [9, Definition 1.13]. Then the proposition follows from [9, Corollary 3.6] and Proposition 2.2. ∎

It is enough to prove that the relation $\leq_{n}$ on $\mathbf{P}(\mathbf{t})$ is transitive. Suppose that $p\leq_{n}q\leq_{n}r$. We want to conclude that $p\leq_{n}r$. Since $q\leq_{n}r$, $\mu_{n}(\mathfrak{g}_{q})=\mu_{n}(\mathfrak{g}_{r})$. Since also $p\leq_{n}q$, we have that for every $i\leq\mu_{n}(\mathfrak{g}_{q})=\mu_{n}(\mathfrak{g}_{r})$,

Therefore, $p\leq_{n}r$. ∎

The only items in Definition 3.1 that require a proof are $\eqref{3.typeBdefin}$ and $\eqref{4.typeBdefin}$:

$\eqref{3.typeBdefin}$ Let $(p_{n}:n\in\mathbf{N})$ be a sequence of conditions such that $p_{n+1}\leq_{n}p_{n}$, and let

If it turns out that $p\in\mathbf{P}(\mathbf{t})$, then $p\leq_{n}p_{n}$ for all $n\in\mathbf{N}$. Thus, it is enough to prove that the gap-counting function $\mathfrak{g}_{p}$ is staggered divergent:

$(3.1)$ To see that $\mathfrak{g}_{p}$ is non-decreasing, let $m\in\mathbf{N}$ be fixed. Since

we can choose some $n\in\mathbf{N}$ such that $m\leq\mu_{n}(\mathfrak{g}_{p_{n}})$. Note that for every $i\leq\mu_{n}(\mathfrak{g}_{p_{n}})$, it holds that $[i]_{\zeta}\setminus\operatorname{\mathrm{dom}}p_{n}=[i]_{\zeta}\setminus\operatorname{\mathrm{dom}}p$. In particular, we have that if $i\leq\mu_{n}(\mathfrak{g}_{p_{n}})$, then $\mathfrak{g}_{p}(i)=\mathfrak{g}_{p_{n}}(i)$. This implies that $\mathfrak{g}_{p}$ is non-decreasing in the interval $[0,m]$. Since this holds for every $m\in\mathbf{N}$, $\mathfrak{g}_{p}$ is non-decreasing.

$(3.2)$ Let us check now that $\mathfrak{g}_{p}$ is divergent. Fix a non-negative integer $n\in\mathbf{N}$. As before, for every $i\leq\mu_{n}(\mathfrak{g}_{p_{n}})$, we have that $\mathfrak{g}_{p}(i)=\mathfrak{g}_{p_{n}}(i)$. Thus,

Since $n\in\mathbf{N}$ was arbitrary, and $\mathfrak{g}_{p}$ is non-decreasing, we have that $\mathfrak{g}_{p}$ is divergent.

$\eqref{4.typeBdefin}$ Let $q\in\mathbf{P}(\mathbf{t})$ be such that $q\Vdash``\tau\in V"$, fix $n\in\mathbf{N}$, and set

Let also $\{s_{k}:k<l\}$ be the finite set of partial $\mathbf{t}$-selectors with domain $F$.

We will define $p\leq_{n}q$ recursively: Start by choosing a condition $q_{0}\leq q$ such that $\operatorname{\mathrm{dom}}q\cup F\subseteq\operatorname{\mathrm{dom}}q_{0}$ and $q_{0}\upharpoonright F=s_{0}$. Then choose some $p_{0}\leq q_{0}$ such that there exists some $x_{0}\in V$ such that $p_{0}\Vdash``\tau=x_{0}"$. Now, if $k\geq 1$, and we already defined $p_{i}$, for $i<k$, let $q_{k}\leq q$ be such that $\operatorname{\mathrm{dom}}q_{k}=\operatorname{\mathrm{dom}}p_{k-1}$, $q_{k}\upharpoonright F=s_{k}$, and $q_{k}\upharpoonright(\operatorname{\mathrm{dom}}q_{k}\setminus F)=p_{k-1}\upharpoonright(\operatorname{\mathrm{dom}}p_{k-1}\setminus F)$. Then choose some $p_{k}\leq q_{k}$ such that there exists some $x_{k}\in V$ such that $p_{k}\Vdash``\tau=x_{k}"$. Finally, let $p:=p_{l-1}\upharpoonright(\operatorname{\mathrm{dom}}p_{l-1}\setminus F)$. Clearly, $p\leq_{n}q$. Now, if $H:=\{x_{k}:k<l\}\in V$, and $r\leq p$ is arbitrary, we can find a further extension $s\leq r$ such that $F\subseteq\operatorname{\mathrm{dom}}s$. If $s\upharpoonright F=s_{k}$, then $s\leq p_{k}$, and therefore $s\Vdash``\tau=x_{k}"$. Thus, $p\Vdash``\tau\in H"$, as required. ∎

Let $(\varepsilon_{n}:n\in\mathbf{N})$ be a sequence of positive real numbers. Without loss of generality, let us assume that $\varepsilon_{n+1}<\varepsilon_{n}$ for all $n\in\mathbf{N}$, and that $\varepsilon_{n}\to 0$ as $n\to\infty$. Let $B_{n}$ be the collection of all open subsets $U$ of $\mathbf{C}$ such that

Let also $\rho\colon\mathbf{N}\to\mathbf{N}$ be a partition such that every $[n]_{\rho}$ is infinite, and let $A_{n}$ be the set of finite unions $U_{m_{n,0}}\cup\dots\cup U_{m_{n,k-1}}$ such that:

1. (1)

   $m_{n,0}<m_{n,1}<\dots<m_{n,k-1}$ are all elements of $[n]_{\rho}$.

2. (2)

   $U_{m_{n,i}}\in B_{m_{n,i}}$ for every $i<k$.

Since the sequence $\varepsilon_{n}\to 0$ as $n\to\infty$, we may assume that no $A_{n}$ has $\mathbf{C}$ as an element, and therefore every $A_{n}$ is an $\omega$-cover of $\mathbf{C}$. Using the selection principle $\mathsf{S}_{1}(\Omega[\mathbf{C}],G\Lambda[X|\mathbf{C}])$, choose open sets $V_{n}\in A_{n}$ such that $\{V_{n}:n\in\mathbf{N}\}\in G\Lambda[X|\mathbf{C}]$. Now, for each $n\in\mathbf{N}$, let $k(n)\in\mathbf{N}$ be such that $V_{n}=U_{m_{n,0}}\cup\dots\cup U_{m_{n,k(n)-1}}$ in a way such that $m_{n,0}<\dots<m_{n,k(n)-1}$ are all elements of $[n]_{\rho}$, and $U_{m_{n,i}}\in B_{m_{n,i}}$ for every $i<k(n)$. Since all the $U_{m_{n,i}}$’s have different diameters (in particular they are different), the set $\left\{U_{m_{n,i}}:n\in\mathbf{N}\,\&\,i<k(n)\right\}$ belongs to $G\Lambda[X|\mathbf{C}]$. The argument finishes by recalling that $\operatorname{\mathrm{diam}}U_{m_{n,i}}\leq\varepsilon_{n}$. ∎

It is enough to prove that $\neg(1)\Rightarrow\neg(2)$. We proceed by cases:

1. (i)

   If $\mathfrak{b}=\aleph_{1}$, it is a result of Bartoszyński (see [1, Theorem 2 (1)]) that there exists an uncountable meager-additive subset of the real line.

2. (ii)

   If $\mathfrak{b}>\aleph_{1}$, let $X$ be a strong measure zero set of reals with $|X|=\aleph_{1}$. Since $\mathfrak{b}>\aleph_{1}$, $X$ has the Hurewicz property (see [12]). By [7, Theorem 8], $X$ is a set of reals that has the Hurewicz property and such that the selection principle $\mathsf{S}_{1}(O[X],O[X])$ holds. By [19, Theorem 14], this implies that the selection principle $\mathsf{S}_{1}(\Omega[\mathbf{R}],G\Lambda[X|\mathbf{R}])$ holds. Therefore, by Proposition 4.2, $X$ has sharp measure zero, and is meager-additive.

This concludes the proof. ∎

Let $M$ be a model of set theory in which all the items in Theorem 4.4 hold. By Proposition 4.2, $\mathscr{N}^{+}(\mathbf{C})=\mathscr{N}^{\star}(\mathbf{C})$. Now, if $X$ is a strong measure zero set of real numbers that it is not meager-additive, since both $\mathscr{N}^{+}(\mathbf{R})$ and $\mathscr{N}^{\star}(\mathbf{R})$ are $\sigma$-ideals (cf. Poposition 2.1), we may assume that $X\subseteq[0,1]$. By Proposition 2.2, $c^{-1}(X)\in\mathscr{N}^{+}(\mathbf{C})\setminus\mathscr{N}^{\star}(\mathbf{C})$; contradiction. An analogous argument proves that, in $M$, there are uncountable strong measure zero sets of the reals. ∎

That all the $D_{n}$’s and all the $E_{x}$’s are open is obvious, so we only need to check the density.

1. (1)

   Let $q\in\mathbf{P}(\hat{\mathbf{T}})$ be a condition, and let $n\in\mathbf{N}$ be fixed. We are looking for a partial $\hat{\mathbf{T}}$-selector $p$ extending $q$, with $n\in\operatorname{\mathrm{dom}}p$, and such that the gap-counting $\mathfrak{g}_{p}$ is still monotone. If $n\in\operatorname{\mathrm{dom}}q$, there is nothing to prove. Otherwise, let $m\in\mathbf{N}$ be such that $n\in[m]_{\zeta}$, and let $\{m-i:i<l\}$ be the set of all the non-negative integers such that $\mathfrak{g}_{q}(m)=\mathfrak{g}_{q}(m-i)$. For each $i<l$, choose $k_{i}\in[m-i]_{\zeta}\setminus\operatorname{\mathrm{dom}}q$, with $k_{0}=n$, and choose arbitrary open sets $U_{k_{i}}\in t_{k_{i}}$. Setting $p:=q\cup\{(k_{i},U_{k_{i}}):i<l\}$, we obtain that $p\in\mathbf{P}(\hat{\mathbf{T}})$ is the desired condition.

2. (2)

   Let $q\in\mathbf{P}(\hat{\mathbf{T}})$, and let $x\in\mathbf{C}$ be fixed. For each $n\geq\mu_{1}(\mathfrak{g}_{q})$, choose some $k_{n}\in[n]_{\zeta}\setminus\operatorname{\mathrm{dom}}q$. Since each $t_{k_{n}}$ is an open cover of $\mathbf{C}$, there must exist open sets $U_{k_{n}}\in t_{k_{n}}$ such that $x\in U_{k_{n}}$ for all $n\in\mathbf{N}$. If we define

   $$p:=q\cup\left\{\left(k_{n},U_{k_{n}}\right):n\geq\mu_{1}(\mathfrak{g}_{q})\right\},$$

   then the condition $p\in\mathbf{P}(\hat{\mathbf{T}})$ is as desired.

This concludes the argument. ∎

Fix a $V$-generic filter $G$ on $\mathbf{P}(\hat{\mathbf{T}})$.

1. (1)

   For each $n\in\mathbf{N}$, let $p\in G\cap D_{n}$. Then $n\in\operatorname{\mathrm{dom}}p\subseteq\operatorname{\mathrm{dom}}\phi_{G}$.

2. (2)

   If $n\in\operatorname{\mathrm{dom}}p$, since $p$ is a partial $\hat{\mathbf{T}}$-selector, $\phi_{G}(n)=p(n)\in t_{n}$.

3. (3)

   For each $x\in\mathbf{C}$, choose a condition $p\in G\cap E_{x}$. Then $x$ is in all but finitely many elements of the set

   $$\left\{\bigcup\left\{p(k):k\in[n]_{\zeta}\cap\operatorname{\mathrm{dom}}p\right\}:n\in\mathbf{N}\right\},$$

   and therefore (3) follows readily.

∎

The model in which we are interested is obtained by doing a countable support iteration of $\mathbf{P}(\hat{\mathbf{T}}_{\alpha})$, for $\alpha<\omega_{2}$, over a model of $V=L$, where at each stage in the iteration, $V[G_{\alpha}]\models``\mathbf{T}_{\alpha}\text{ is a sequence of }\omega\text{-covers of }\mathbf{C}"$, and where we have dovetailed so as to ensure that for any $\mathbf{T}$ such that

then for cofinally many $\alpha<\omega_{2}$ we have that $\mathbf{T}=\mathbf{T}_{\alpha}$. This dovetailing can be done since there are only continuum many sequences of $\omega$-covers of $\mathbf{C}$, and the intermediate models satisfy the continuum hypothesis.

We may continue now with the proof of Theorem 4.4:

1. (1)

   By Proposition 3.4, every strong measure zero subset $X$ of $\mathbf{C}$ in $V[G_{\omega_{2}}]$ has cardinality at most $\aleph_{1}$, so it is $\zeta$-supernull.

2. (2)

   By Corollary 3.3, every function $f\colon\mathbf{N}\to\mathbf{N}$ in $V[G_{\omega_{2}}]$ is dominated by some function $g\colon\mathbf{N}\to\mathbf{N}$ in the ground model.

3. (3)

   Working in $V[G_{\omega_{2}}]$: By Proposition 3.4, $\mathscr{N}^{+}(\mathbf{C})\subseteq[\mathbf{C}]^{\leq\aleph_{1}}$. On the other hand, if $X$ is a subset of $\mathbf{C}$ with $|X|\leq\aleph_{1}$, then $X$ is $\zeta$-supernull. By Proposition 4.2, $X$ has sharp measure zero, and by Proposition 2.1, $X$ has strong measure zero as well.

4. (4)

   Follows from the usual argument.

∎