Mobile families of sets and pure subsets Exercise
This is Bourbaki Theory of Sets Chapter 3 Section 4 Exercise 11, English version:
Let $A$ be a set and let $\mathcal{R}$ be a subset of the set $\mathcal{F}(A)$ of finite subsets of $A$. $\mathcal{R}$ is said to be mobile if it satisfies the following condition:
(MO) If $X$, $Y$ are two distinct elements of $\mathcal{R}$ and if $z\in X\cap Y$, then there exists $Z\subseteq X\cap Y$ belonging to $\mathcal{R}$ such that $z\notin Z$.
A subset $P$ of $A$ is then said to be pure if it contains no set belonging to $\mathcal{R}$.
a) Show that every pure subset of $A$ is contained in a maximal pure subset of $A$.
b) Let $M$ ba a maximal pure subset of $A$. Show that for each $x\in X\setminus M$ there exists a unique finite subset $E_{M}(x)$ of $M$ such that $E_{M}(x)\cup\{x\}\in\mathcal{R}$. Moreover, if $y\in E_M(x)$, the set $(M\cup\{x\})\setminus\{y\}$ is a maximal pure subset of $A$.
c) Let $M$, $N$ be two maximal pure subsets of $A$, such that $N\setminus M$ is finite. Show that $|M|=|N|$.
d) Let $M$, $N$ be two maximal pure subsets of $A$, and put $N'=N\setminus M$, $M'=M\setminus N$. Show that:
$$M'\subseteq\bigcup_{x\in N'}E_M(x).$$
Deduce that $|M|=|N|$.
Question
I was able to do all this exercise, but, in the French version, the exercise assumes only a weaker condition:
(MO') If $X$, $Y$ are two distinct elements of $\mathcal{R}$ and if $z\in X\cap Y$, then there exists $Z\subseteq X\cup Y$ belonging to $\mathcal{R}$ such that $z\notin Z$
instead of (MO), and was able to do items (a) to (c), but I do not know how to do item (d).
Attempt
There is my solution to items (a) to (c) assuming only (MO') and item (d) assuming (MO).
a) Straightforward application of Zorn's lemma. In fact, if $\mathcal{A}$ is the set of all pure subsets of $A$ containing a given pure subset $P$, then for every totally ordered subset $\mathcal{C}$ of $\mathcal{A}$, if $\bigcup\mathcal{C}$ is not pure, then it contains an $R\in\mathcal{R}$, but $R$ is finite, say, $R=\{a_1,\dots,a_n\}$, so for $i=1,\dots,n$ there is a $Q_i\in\mathcal{C}$ such that $a_i\in Q_i$, then there is a $Q\in\mathcal{C}$ such that $Q_1,\dots,Q_n\subseteq Q$, so $R\subseteq Q$, contradicting the purity of $Q$; therefore $\bigcup\mathcal{C}\in\mathcal{P}$.
b) Because of maximality of $M$, there is an $E\in\mathcal{F}(M)$ such that $E\cup\{x\}\in\mathcal{R}$.
If $F\in\mathcal{F}(M)$, $F\neq E$ and $F\cup\{x\}\in\mathcal{R}$, then $E\cup\{x\}\neq F\cup\{x\}$ and $x\in\left(E\cup\{x\}\right)\cap\left(F\cup\{x\}\right)$, so by (MO') the set $\left(\left(E\cup\{x\}\right)\cup\left(F\cup\{x\}\right)\right)\setminus\{x\}$ is not pure, but:
$$\left(\left(E\cup\{x\}\right)\cup\left(F\cup\{x\}\right)\right)\setminus\{x\}\subseteq E\cup F\subseteq M,$$
so $M$ will not be pure, a contradiction.
Let $y\in E$. If $R\subseteq(M\cup\{x\})\setminus\{y\}$ and $R\in\mathcal{R}$, then $R\subseteq M\cup\{x\}$, so, by purity of $M$, $R=G\cup\{x\}$ for some $G\in\mathcal{F}(M)$, and by (a) we have $G=E$, but $y\notin G$ and $y\in E$, a contradiction. Therefore $(M\cup\{x\})\setminus\{y\}$ is pure.
Let $z\in A\setminus((M\cup\{x\})\setminus\{y\})$, then $z=y\text{ or }(z\notin M\text{ and }z\neq x)$.
For the case $z=y$, we have $((M\cup\{x\})\setminus\{y\})\cup\{y\}=M\cup\{x\}$, that is not pure.
For the case $z\neq y$, then $z\notin M$ and $z\neq x$, so there is a $H\in\mathcal{F}(M)$ such that $H\cup\{z\}\in\mathcal{R}$, so:

*

*If $y\notin H\cup\{z\}$, then:

$$H\cup\{z\}\subseteq (M\cup\{z\})\setminus\{y\}\subseteq((M\cup\{x\})\setminus\{y\})\cup\{z\}.$$


*If $y\in S$, then $y\in(E\cup\{x\})\cap(H\cup\{z\})$, so $((E\cup\{x\})\cup(H\cup\{z\}))\setminus\{y\}$ is not pure, and:

$$((E\cup\{x\})\cup(H\cup\{z\}))\setminus\{y\}\subseteq((M\cup\{x\})\setminus\{y\})\cup\{z\}.$$
Therefore $((M\cup\{x\})\setminus\{y\})\cup\{z\}$ is not pure. So $(M\cup\{x\})\setminus\{y\}$ is maximal.
c) Induction on $|N\setminus M|$.

*

*If $|N\setminus M|=0$, then $N\subseteq M$, so by maximality of $N$ we have $N=M$, so $|M|=|N|$.


*If $|N\setminus M|>0$, then there is a $m\in M\setminus N$ and there is an $n\in E_N(m)$, so by item (b) the set $N'=(N\cup\{m\})\setminus\{n\}$ is pure maximal and $|N'\setminus M|<|N\setminus M|$, so by induction hypothesis we have $|M|=|N'|$, but $|N'|=|N|$, so $|M|=|N|$.
d) If we assume (MO), then for $m\in M'$ we have $E_N(m)\cup\{m\}\in\mathcal{R}$, so $E_N(m)\cup\{m\}\nsubseteq M$, but $m\in M$, so $E_N(m)\nsubseteq M$, so there is an $x\in E_N(m)$ such that $x\notin M$, so $x\in N'$, and $E_M(x)\cup\{x\}\in\mathcal{R}$, therefore:
$$x\in(E_M(x)\cup\{x\})\cap(E_N(m)\cup\{m\}),$$
so we have two cases:

*

*If $E_M(x)\cup\{x\}=E_N(m)\cup\{m\}$, then $x\in N'$ and $m\in E_M(x)$.


*If $E_M(x)\cup\{x\}\neq E_N(m)\cup\{m\},$ then by (MO) the set $((E_M(x)\cup\{x\})\cap(E_N(m)\cup\{m\}))\setminus\{x\}$ is not pure, but it is contained in $M$, a contradiction.
Finally, by virtue of (c), we are reduced to the case where $M'$ and $N'$ are infinite, so:
$$|M'|\leq|\bigcup_{x\in N'}E_M(x)|\leq\sum_{x\in N'}|E_M(x)|\leq\sum_{x\in N'}\aleph_0=|N'|\aleph_0=|N'|,$$
and analogously $|N'|\leq |M'|$, so $|M'|=|N'|$, and we conclude that $|M|=|N|$.
 A: I don't know how to prove $M'\subseteq\bigcup_{x\in N'}E_M(x)$, but here is a proof that $|M|=|N|$.  Suppose $|M|\neq |N|$; we may assume $|M|>|N|$ and that $M$ and $N$ are infinite.  Call a finite subset $E\subset N$ bad if there exists an infinite family $S\subseteq\mathcal{F}(M)$ such that the elements of $S$ are pairwise disjoint and $E\cup F\in\mathcal{R}$ for all $F\in S$.
I first claim that a bad set exists.  To prove this, note that since $|M|>|N|$ we have $|M'|=|M|>|N|=|\mathcal{F}(N)|$.  Thus there is an infinite subset $M_0\subseteq M'$ such that $E_N(x)=E_N(y)$ for all $x,y\in M_0$.  Then the common value of $E_N(x)$ for $x\in M_0$ is bad, because we can take $S=\{\{x\}:x\in M_0\}$.
Since a bad set exists, there exists a minimal bad set $E$, i.e. a bad set $E$ such that no proper subset is bad.  Let $S\subseteq\mathcal{F}(M)$ witness that $E$ is bad.  Note that the empty set is not bad since $M$ is pure, so $E$ is nonempty; pick an element $x\in E$.  Let $P$ be a partition of $S$ into pairs.  For each $\{F,G\}\in P$, both $E\cup F$ and $E\cup G$ are in $\mathcal{R}$, so by (MO'), we can pick some $H(\{F,G\})\subseteq (E\cup F\cup G)\setminus\{x\}$ which is in $\mathcal{R}$.  Let $E(\{F,G\})=H(\{F,G\})\cap E$ and $I(\{F,G\})=H(\{F,G\})\cap (F\cup G)$.  By Pigeonhole, there is some subset $E'\subseteq E$ such that $E(\{F,G\})=E'$ for infinitely many $\{F,G\}\in P$.  Since the elements of $S$ are disjoint, so are the sets $I(\{F,G\})$, and moreover each $I(\{F,G\})$ is nonempty since $N$ is pure.  So $S'=\{I(\{F,G\}):E(\{F,G\})=E'\}$ is an infinite set of disjoint finite subsets of $M$.  Moreover, for each $I(\{F,G\})\in S'$, $E'\cup I(\{F,G\})=H(\{F,G\})$ is in $\mathcal{R}$.  Thus $E'$ is bad.  But $x\not\in E'$, so $E'$ is a proper subset of $E$, contradicting the minimality of $E$.
A: Let me start with some context. Minimal sets in $\mathcal{R}$ are exactly the circuits of a (finitary) matroid on $A$, i.e. minimal dependent sets. A matroid is a notion that generalizes linear dependence in a vector space. In particular, pure maximal sets are exactly the bases of said matroid. The invariance of their cardinal is thus a well known fact that generalizes the invariance of the cardinal of bases in vector spaces, cf. Rado, R. 1949. « Axiomatic Treatment of Rank in Infinite Sets ». Canadian Journal of Mathematics 1 (4): 337‑43. https://doi.org/10.4153/CJM-1949-031-1.
However, that does not answer the first part of question d. Knowing that we are looking at a matroid also simplifies that question. Let $Y = \bigcup_{x\in N'} E_M(x)$. Then $Y \subset M$ generates N'. It follows that $Y \cup (M\cap N)$ generates $N$ and hence is a subset of $M$ which is itself a basis. So it is equal to $M$ and we do have $M' \subset Y$.
That being said, one can prove the inclusion directly. A complete solution (in French) to the exercice can be found here: https://www.bourbaki.fr/TEXTES/E-III-4-11.pdf. Let me translate the argument for the proof of $M' \subset \bigcup_{x\in N'} E_M(x)$.
Fix $y\in M'$. Let us prove that there exists $x \in E_N(y)\setminus M$ such that $y\in E_M(x)$. We proceed by induction on the cardinal of $E_N(y)\setminus M$, which is non-empty, by purity of $M$. Pick some $x \in E_N(y)\setminus M$. We also have that $E_M(x)\setminus N$ is non-empty and pick $y'$ in it. If $y = y'$ we are done. Otherwise, consider $M'' = (M\cup\{x\})\setminus \{y\}$ which is maximal pure. Since $E_{N}(y)\cap M'' \subset (E_N(y)\cap M)\setminus \{x\}$, by induction, we find $x'\in E_N(y)\cap M''$ such that $y\in E_{M''}(x)$.
If $y' \not\in E_{M}(x')$, then $E_M(x') \subset M''$ and hence $E_M(x') = E_{M''}(x')$ contains $y$. Otherwise, by (MO'), there exists some $Z\subseteq E_M(x)\cup\{x\}\cup E_M(x')\cup\{x'\}$ with $y'\not\in Z$. Then $Z\subset M''\cup\{x'\}$ and hence is equal to $E_{M''}(x')$; a set that contains $y$. It follows that either $E_M(x)$ or $E_M(x')$ contains $y$.
