[proofplan] Throughout the proof, $\omega$ denotes the set of natural numbers, $\omega^\omega$ denotes the set of all maps from $\omega$ to $\omega$, $\mathbb{R}$ denotes the real line with its usual topology and [Lebesgue measure](/page/Lebesgue%20Measure), and $\mathfrak{c}:=|\mathbb{R}|$ denotes the cardinality of the continuum. We use three standard consequences of Martin's axiom for ccc partial orders. First, every family of fewer than $\mathfrak{c}$ functions in $\omega^\omega$ has a common eventual dominator, which gives $\mathfrak{b}=\mathfrak{d}=\mathfrak{c}$. Second, the Martin's axiom category theorem says that the union of fewer than $\mathfrak{c}$ meagre subsets of a Polish space is meagre; applying this to $\mathbb{R}$ gives the meagre ideal equalities. Third, the Martin's axiom measure theorem says that the union of fewer than $\mathfrak{c}$ Lebesgue null subsets of $\mathbb{R}$ is null; this gives $\operatorname{add}(\mathcal{N})=\operatorname{cov}(\mathcal{N})=\mathfrak{c}$. The additional hypothesis $\neg\mathrm{CH}$ is not used in the derivation of these equalities; it only records that the application is in the usual non-continuum-hypothesis context where $\mathfrak{c}>\aleph_1$. [/proofplan]

[step:Dominate every small family in $\omega^\omega$]Let $\omega^\omega$ denote the set of maps $f: \omega \to \omega$, ordered by eventual domination: for $f,g \in \omega^\omega$, write $f \leq^* g$ if there exists $N \in \omega$ such that $f(n) \leq g(n)$ for every $n \geq N$. Let $F \subset \omega^\omega$ satisfy $|F| < \mathfrak{c}$. We prove that $F$ is bounded in $(\omega^\omega,\leq^*)$. Define a partial order $\mathbb{P}$ as follows. A condition is a pair $(s,A)$ where $s: m \to \omega$ is a finite sequence for some $m \in \omega$ and $A \subset F$ is finite. Put $(t,B) \leq (s,A)$ if $s \subset t$, $A \subset B$, and for every $f \in A$ and every $n \in \operatorname{dom}(t) \setminus \operatorname{dom}(s)$ one has $t(n) \geq f(n)$. The order is ccc because, for each fixed finite sequence $s$, any two conditions with first coordinate $s$ are compatible, and there are only countably many finite sequences $s: m \to \omega$. For each $k \in \omega$, define \begin{align*} D_k := \{(s,A) \in \mathbb{P} : k \in \operatorname{dom}(s)\}. \end{align*} This set is dense: given $(s,A)$, extend $s$ to a sequence whose domain contains $k$, choosing new values at coordinates $n$ at least $\max\{f(n): f \in A\}$, with the maximum over the empty set interpreted as $0$. For each $f \in F$, define \begin{align*} E_f := \{(s,A) \in \mathbb{P} : f \in A\}. \end{align*} This set is dense because $(s,A \cup \{f\}) \leq (s,A)$. The family of dense sets $\{D_k : k \in \omega\} \cup \{E_f : f \in F\}$ has cardinality $|F|+\aleph_0<\mathfrak{c}$. By Martin's axiom for ccc partial orders, there is a filter $G \subset \mathbb{P}$ meeting all these dense sets. Define $g: \omega \to \omega$ by declaring $g(k)=s(k)$ whenever $(s,A) \in G$ and $k \in \operatorname{dom}(s)$. This is well-defined because any two conditions in the filter have a common extension, and extensions preserve finite sequence values. Since $G$ meets every $D_k$, the value $g(k)$ is defined for every $k \in \omega$. Fix $f \in F$. Since $G$ meets $E_f$, choose $(s,A) \in G$ with $f \in A$. If $n \geq |s|$, then choose $(t,B) \in G$ with $n \in \operatorname{dom}(t)$ and take a common extension $(u,C) \in G$ of $(s,A)$ and $(t,B)$. The definition of the order gives $u(n) \geq f(n)$, and $u(n)=g(n)$. Hence $f \leq^* g$. Therefore every subset of $\omega^\omega$ of cardinality less than $\mathfrak{c}$ is bounded.[/step]

[guided]The goal of this step is to turn Martin's axiom into a single function that eventually dominates a prescribed small family. Let $F \subset \omega^\omega$ be a family with $|F|<\mathfrak{c}$. We use forcing only as an organizing device: a condition records a finite initial segment of the future dominating function and a finite list of functions from $F$ that must be dominated from that point onward. A condition is a pair $(s,A)$ where $s:m\to\omega$ is a finite sequence and $A\subset F$ is finite. The extension relation is designed so that once a function $f$ has entered the finite side condition $A$, all newly chosen coordinates must dominate $f$: $(t,B)\leq(s,A)$ means $s\subset t$, $A\subset B$, and $t(n)\geq f(n)$ for every $f\in A$ and every new coordinate $n\in\operatorname{dom}(t)\setminus\operatorname{dom}(s)$. We verify that $\mathbb{P}$ is ccc. Fix a finite sequence $s$. If $(s,A)$ and $(s,B)$ are two conditions with the same first coordinate, then $(s,A\cup B)$ extends both of them. Thus every antichain contains at most one condition above each finite sequence $s$. Since there are only countably many finite sequences of natural numbers, every antichain in $\mathbb{P}$ is countable. This is exactly the countable chain condition. Now define the dense sets that force the generic object to be a total function and to dominate every member of $F$. For $k\in\omega$, let \begin{align*} D_k := \{(s,A)\in\mathbb{P}: k\in\operatorname{dom}(s)\}. \end{align*} Given any condition $(s,A)$, we extend $s$ until coordinate $k$ is included. At each new coordinate $n$, choose a value at least $\max\{f(n):f\in A\}$, using $0$ if $A=\varnothing$. This produces a stronger condition in $D_k$, so $D_k$ is dense. For $f\in F$, let \begin{align*} E_f := \{(s,A)\in\mathbb{P}: f\in A\}. \end{align*} This is dense because adding $f$ to the finite side condition gives $(s,A\cup\{f\})\leq(s,A)$. The dense family has size $|F|+\aleph_0$, which is less than $\mathfrak{c}$. Since $\mathbb{P}$ is ccc, Martin's axiom gives a filter $G\subset\mathbb{P}$ meeting every $D_k$ and every $E_f$. Define $g:\omega\to\omega$ by setting $g(k)=s(k)$ whenever $(s,A)\in G$ and $k\in\operatorname{dom}(s)$. This definition is independent of the chosen condition: two conditions in a filter have a common extension, and extensions do not change previously assigned finite sequence values. The filter meets every $D_k$, so $g$ is total. Finally fix $f\in F$. Since $G$ meets $E_f$, choose $(s,A)\in G$ with $f\in A$. For any $n\geq |s|$, choose a condition in $G$ whose finite sequence includes $n$, and then choose a common extension $(u,C)\in G$ of that condition and $(s,A)$. Because $f\in A$ and $n$ is a coordinate added after $s$, the order relation forces $u(n)\geq f(n)$. Since $u$ is an initial segment of the function $g$, this says $g(n)\geq f(n)$. Thus $f\leq^* g$. As $f$ was arbitrary, $g$ dominates all of $F$.[/guided]

[step:Conclude $\mathfrak{b}=\mathfrak{d}=\mathfrak{c}$] By the previous step, no family in $\omega^\omega$ of cardinality less than $\mathfrak{c}$ is unbounded. Hence $\mathfrak{b}\geq\mathfrak{c}$. Since $\omega^\omega$ has cardinality $\mathfrak{c}$ and is unbounded in itself, $\mathfrak{b}\leq\mathfrak{c}$. Therefore $\mathfrak{b}=\mathfrak{c}$. The inequalities $\mathfrak{b}\leq\mathfrak{d}\leq\mathfrak{c}$ hold by definition: every dominating family is unbounded, and $\omega^\omega$ itself is a dominating family of size $\mathfrak{c}$. Combining these inequalities with $\mathfrak{b}=\mathfrak{c}$ gives $\mathfrak{d}=\mathfrak{c}$. [/step]

[step:Apply the Martin's axiom category theorem to the meagre ideal]Let $\mathcal{M}$ be the ideal of meagre subsets of $\mathbb{R}$. We invoke the Martin's axiom category theorem, a standard consequence of Martin's axiom for ccc partial orders: if $X$ is a Polish space and $\{A_i:i\in I\}$ is a family of meagre subsets of $X$ with $|I|<\mathfrak{c}$, then $\bigcup_{i\in I}A_i$ is meagre in $X$. The hypotheses are satisfied because $\mathbb{R}$ with its usual topology is a Polish space and the family under consideration has cardinality less than $\mathfrak{c}$. Applying this theorem to $X=\mathbb{R}$ gives that every union of fewer than $\mathfrak{c}$ meagre subsets of $\mathbb{R}$ is meagre. It follows that $\operatorname{add}(\mathcal{M})\geq\mathfrak{c}$, because no subfamily of $\mathcal{M}$ of size less than $\mathfrak{c}$ has nonmeagre union. Since there are exactly $\mathfrak{c}$ singleton subsets of $\mathbb{R}$ and their union is $\mathbb{R}$, which is nonmeagre, we also have $\operatorname{add}(\mathcal{M})\leq\mathfrak{c}$. Hence $\operatorname{add}(\mathcal{M})=\mathfrak{c}$. The same category theorem gives $\operatorname{cov}(\mathcal{M})\geq\mathfrak{c}$, because fewer than $\mathfrak{c}$ meagre sets cannot cover $\mathbb{R}$. The cover of $\mathbb{R}$ by its singleton subsets has size $\mathfrak{c}$, so $\operatorname{cov}(\mathcal{M})\leq\mathfrak{c}$. Thus $\operatorname{cov}(\mathcal{M})=\mathfrak{c}$.[/step]

[guided]The category input is the Martin's axiom category theorem. It says that under Martin's axiom for ccc partial orders, whenever $X$ is a Polish space and $\{A_i:i\in I\}$ is a family of meagre subsets of $X$ indexed by a set $I$ with $|I|<\mathfrak{c}$, the union $\bigcup_{i\in I}A_i$ is still meagre in $X$. We apply it with $X=\mathbb{R}$. The space $\mathbb{R}$ with the usual metric is Polish, and the families used in the definitions of $\operatorname{add}(\mathcal{M})$ and $\operatorname{cov}(\mathcal{M})$ are exactly families of meagre subsets of $\mathbb{R}$. For additivity, the theorem says that no family of fewer than $\mathfrak{c}$ members of $\mathcal{M}$ can have a nonmeagre union. Therefore $\operatorname{add}(\mathcal{M})\geq\mathfrak{c}$. On the other hand, each singleton subset of $\mathbb{R}$ is nowhere dense, hence meagre, and $\mathbb{R}$ is the union of its $\mathfrak{c}$ singleton subsets. Since $\mathbb{R}$ is not meagre in itself by the [Baire category theorem](/theorems/630), this family witnesses $\operatorname{add}(\mathcal{M})\leq\mathfrak{c}$. Thus $\operatorname{add}(\mathcal{M})=\mathfrak{c}$. For covering, if fewer than $\mathfrak{c}$ meagre sets covered $\mathbb{R}$, then their union would be all of $\mathbb{R}$ and hence nonmeagre, contradicting the Martin's axiom category theorem. Thus $\operatorname{cov}(\mathcal{M})\geq\mathfrak{c}$. The singleton cover of $\mathbb{R}$ has size $\mathfrak{c}$ and consists of meagre sets, so $\operatorname{cov}(\mathcal{M})\leq\mathfrak{c}$. Hence $\operatorname{cov}(\mathcal{M})=\mathfrak{c}$.[/guided]

[step:Derive $\operatorname{non}(\mathcal{M})=\operatorname{cof}(\mathcal{M})=\mathfrak{c}$] Because $\operatorname{add}(\mathcal{M})=\mathfrak{c}$, every subset of $\mathbb{R}$ of cardinality less than $\mathfrak{c}$ is meagre: it is a union of fewer than $\mathfrak{c}$ singleton meagre sets. Hence no nonmeagre subset of $\mathbb{R}$ has cardinality less than $\mathfrak{c}$, so $\operatorname{non}(\mathcal{M})\geq\mathfrak{c}$. Since $\mathbb{R}$ itself is nonmeagre and has cardinality $\mathfrak{c}$, $\operatorname{non}(\mathcal{M})\leq\mathfrak{c}$. Therefore $\operatorname{non}(\mathcal{M})=\mathfrak{c}$. For the cofinality, first note that $\operatorname{cof}(\mathcal{M})\leq\mathfrak{c}$ because every meagre subset of $\mathbb{R}$ is contained in a Borel meagre subset of $\mathbb{R}$, and there are only $\mathfrak{c}$ Borel subsets of $\mathbb{R}$. Conversely, if $\mathcal{B}\subset\mathcal{M}$ is cofinal in $\mathcal{M}$, then $\bigcup\mathcal{B}=\mathbb{R}$ because each singleton belongs to some member of $\mathcal{B}$. Therefore $|\mathcal{B}|\geq\operatorname{cov}(\mathcal{M})=\mathfrak{c}$. Thus $\operatorname{cof}(\mathcal{M})\geq\mathfrak{c}$, and consequently $\operatorname{cof}(\mathcal{M})=\mathfrak{c}$. [/step]

[step:Apply the Martin's axiom measure theorem to the null ideal]Let $\mathcal{N}$ be the ideal of Lebesgue null subsets of $\mathbb{R}$. We invoke the Martin's axiom measure theorem, a standard consequence of Martin's axiom for ccc partial orders: every union of fewer than $\mathfrak{c}$ Lebesgue null subsets of $\mathbb{R}$ is Lebesgue null. The hypotheses are satisfied because the sets under consideration are Lebesgue null subsets of $\mathbb{R}$ and the indexing families have cardinality less than $\mathfrak{c}$. Therefore no family of fewer than $\mathfrak{c}$ null sets has non-null union, so $\operatorname{add}(\mathcal{N})\geq\mathfrak{c}$. Since $\mathbb{R}$ is the union of its $\mathfrak{c}$ singleton subsets and is not null, $\operatorname{add}(\mathcal{N})\leq\mathfrak{c}$. Hence $\operatorname{add}(\mathcal{N})=\mathfrak{c}$. The same measure theorem implies that fewer than $\mathfrak{c}$ null sets cannot cover $\mathbb{R}$, since their union would be null while $\mathbb{R}$ is not null. Thus $\operatorname{cov}(\mathcal{N})\geq\mathfrak{c}$. The cover of $\mathbb{R}$ by singleton null sets has size $\mathfrak{c}$, so $\operatorname{cov}(\mathcal{N})\leq\mathfrak{c}$. Therefore $\operatorname{cov}(\mathcal{N})=\mathfrak{c}$.[/step]

[guided]The measure input is the Martin's axiom measure theorem. Under Martin's axiom for ccc partial orders, the union of any family of fewer than $\mathfrak{c}$ Lebesgue null subsets of $\mathbb{R}$ is again Lebesgue null. This applies directly to the families appearing in $\operatorname{add}(\mathcal{N})$ and $\operatorname{cov}(\mathcal{N})$, because those families consist of Lebesgue null subsets of $\mathbb{R}$ and, in the lower-bound arguments, have cardinality less than $\mathfrak{c}$. For additivity, if fewer than $\mathfrak{c}$ null sets had non-null union, this would contradict the Martin's axiom measure theorem. Hence $\operatorname{add}(\mathcal{N})\geq\mathfrak{c}$. Conversely, every singleton subset of $\mathbb{R}$ has Lebesgue measure zero, and the union of all singleton subsets of $\mathbb{R}$ is $\mathbb{R}$ itself. Since $\mathbb{R}$ is not Lebesgue null and there are exactly $\mathfrak{c}$ singleton subsets, we get $\operatorname{add}(\mathcal{N})\leq\mathfrak{c}$. Therefore $\operatorname{add}(\mathcal{N})=\mathfrak{c}$. For covering, if fewer than $\mathfrak{c}$ null sets covered $\mathbb{R}$, their union would be $\mathbb{R}$. The Martin's axiom measure theorem would then make $\mathbb{R}$ null, which is false. Thus $\operatorname{cov}(\mathcal{N})\geq\mathfrak{c}$. The singleton cover of $\mathbb{R}$ has size $\mathfrak{c}$ and consists of null sets, so $\operatorname{cov}(\mathcal{N})\leq\mathfrak{c}$. Hence $\operatorname{cov}(\mathcal{N})=\mathfrak{c}$.[/guided]

[step:Combine the computed equalities] The preceding steps show \begin{align*} \mathfrak{b}=\mathfrak{d}=\mathfrak{c}, \end{align*} \begin{align*} \operatorname{add}(\mathcal{M})=\operatorname{cov}(\mathcal{M})=\operatorname{non}(\mathcal{M})=\operatorname{cof}(\mathcal{M})=\mathfrak{c}, \end{align*} and \begin{align*} \operatorname{add}(\mathcal{N})=\operatorname{cov}(\mathcal{N})=\mathfrak{c}. \end{align*} These are exactly the asserted conclusions. [/step]