# Set theory - Chapter 7: Filters, Ultrafilters and Boolean Algebras

## 7.1

Let $F$ be a filter and $X\in F$.

1. (1)

$\emptyset\notin{\operatorname{\mathcal{P}}(X)\cap F}$ (because $F$ is a filter) and $X\in{\operatorname{\mathcal{P}}(X)\cap F}$.

2. (2)

If $A,B\in{\operatorname{\mathcal{P}}(X)\cap F}$, then $A\cap B\subseteq X$ and $A\cap B\in F$ (because $F$ is a filter). So $A\cap B\in{\operatorname{\mathcal{P}}(X)\cap F}$.

3. (3)

Let $A,B\subseteq X$ and $A\in{\operatorname{\mathcal{P}}(X)\cap F}$. If $A\subseteq B$ then $B\in F$ (because $F$ is a filter) so $B\in{\operatorname{\mathcal{P}}(X)\cap F}$.

So $\operatorname{\mathcal{P}}(X)\cap F$ is a filter on $X$.

## 7.2

Let $A$ be an infinite set and $S={[A]}^{{<\omega}}$ be the set of all finite subsets of $A$. For each $P\in S$, let $\hat{P}=\{Q\in S:P\subseteq Q\}$. Let $F$ be the set of all $X\subseteq S$ such that $X\subseteq\hat{P}$ for some $P\in S$. Clearly, $F$ is a filter: we have $P\in\hat{P}$ so $X\subseteq\hat{P}\implies X\neq\emptyset$ ; $\hat{\emptyset}=S$ ; if $X\supseteq\hat{P}$ and $Y\supseteq\hat{Q}$ then $X\cap Y\subseteq\hat{P}\cap\hat{Q}=\hat{P\cup Q}$ ; if $Y\supseteq X$ and $X\supseteq\hat{P}$ then $Y\supseteq\hat{P}$. If there exists some $X_{0}\subseteq S$ such that $F=\{X\subseteq S:X\supseteq X_{0}\}$ then there is $P\in S$ such that $X_{0}=\hat{P}$. Because $P$ is finite and $A$ is infinite, we can find $x\in A\setminus P$ then $\hat{\{a\}}\in F$ but $\hat{\{a\}}\nsupseteq\hat{P}$ ($P\in\hat{P}$ and $P\notin\hat{\{a\}}$) Hence $P$ is nonprincipal.

Let $G=\{\{a\}:a\in A\}$. $G$ has the finite intersection property: for any $a_{1},a_{2},...a_{n}\in A$, $\bigcap_{{i. Let $H$ be the filter generated by $G$. He wave $G\subseteq F$ so $H\subseteq F$. Moreover, if $X\in F$, there is $P\in S$ such that $X\supseteq\hat{P}$. If $P=\emptyset$ then $X=S\in H$. Otherwise, there is $n>0$ and $a_{1},a_{2},...a_{n}\in A$ such that $P=\{a_{1},a_{2},...,a_{n}\}$. We just saw that $\hat{P}=\hat{\{a_{1},a_{2},...,a_{n}\}}\in H$, hence $X\in H$. Finally, $F=H$.

## 7.3

Let $U$ be an ultrafilter on $S$ and $X\cup Y\in U$. If $X,Y\notin U$ then $(S\backslash X),(S\backslash Y)\in U$ and so $S\backslash(X\cup Y)=(S\backslash X)\cap(S\backslash Y)\in U$. A contradiction.

## 7.4

Let $U$ be an ultrafilter on a set $S$. And $V$ the set of all $X\subseteq S\times S$ such that $\hat{X}=\{a\in S:{\{b\in S:(a,b)\in X\}}\in U\}\in U$. We have $\hat{\emptyset}=\{a\in S:\emptyset\in U\}=\emptyset\notin U$ so $\emptyset\notin V$. $\hat{S\times S}=\{a\in S:S\in U\}=S\in U$ so $S\times S\in V$. Let $X,Y\in V$ and $a\in S$ such that $a\in\hat{X}\cap\hat{Y}$ i.e. ${\{b\in S:(a,b)\in X\}}\in U$ and ${\{b\in S:(a,b)\in Y\}}\in U$. Then ${\{b\in S:(a,b)\in{X\cap Y}\}}$ is the intersection of these two sets and so is in $U$. Hence $\hat{X\cap Y}\supseteq\hat{X}\cap\hat{Y}$ and because $\hat{X},\hat{Y}\in U$ we have $\hat{X\cap Y}\in U$. Hence $X\cap Y\in V$. Finally, let $X,Y\in S\times S$ such that $X\in V$ and $Y\supseteq X$. Given $a\in S$, we have ${\{b\in S:(a,b)\in X\}}\subseteq{\{b\in S:(a,b)\in Y\}}$ so if the first one is in $U$ so is the second one. This means that $\hat{X}\subseteq\hat{Y}$. Because $\hat{X}\in U$, we have $\hat{Y}\in U$ and so $Y\in V$. Hence $V$ is a filter on $S\times S$ and it remains to prove it is actually an ultrafilter. Let $X\subset S\times S$ such that $X\notin V$. Then $\hat{X}\notin U$ and so $S\setminus{\hat{X}}\in U$. But ${S\setminus\hat{X}}=\{a\in S:{\{b\in S:(a,b)\in X\}}\notin U\}=\{a\in S:{\{b% \in S:(a,b)\notin X\}}\in U\}=\hat{{S\times S}\setminus X}$. Hence ${{S\times S}\setminus X}\in V$.

## 7.5

Let $U$ be an ultrafilter on $S$ and $f:S\rightarrow T$. Define $f_{\star}(U)=\{X\subseteq T:f_{{-1}}(X)\in U\}$. We have $f_{{-1}}(\emptyset)=\emptyset\notin U$ and $f_{{-1}}(T)=S\in U$ so $\emptyset\notin f_{\star}(U)$ and $T\in f_{\star}(U)$. If $X,Y\in f_{\star}(U)$ then $f_{{-1}}(X\cap Y)=f_{{-1}}(X)\cap f_{{-1}}(Y)\in U$ so $X\cap Y\in f_{\star}(U)$. If $X\in f_{\star}(U)$ and $X\subseteq Y$ then $f_{{-1}}(Y)\supseteq f_{{-1}}(X)\in U$ and so $f_{{-1}}(Y)\in U$ and $Y\in f_{\star}(U)$. Finally, for any $X\subseteq T$ we have $S=f_{{-1}}(T)=f_{{-1}}({X\cup(T\setminus X)})=f_{{-1}}(X)\cup f_{{-1}}(T% \setminus X)$ so by exercise 7.3 either $f_{{-1}}(X)$ or $f_{{-1}}(T\setminus X)$ is in $U$ i.e. $X\in f_{\star}(U)$ or $T\setminus X\in f_{\star}(U)$. Finally $f_{\star}(U)$ is an ultrafilter on $T$.

## 7.6

Suppose $a,b$ are two $U$-limits. Then given any $\epsilon>0$, the sets $\{n:\left|a_{n}-a\right|<\frac{\epsilon}{2}\}$ and $\{n:\left|a_{n}-b\right|<\frac{\epsilon}{2}\}$ are in $U$. As a consequence, the intersection of these sets is in $U$ and in particular non empty: there exists $n$ such that $\left|a_{n}-a\right|<\frac{\epsilon}{2}$ and $\left|a_{n}-b\right|<\frac{\epsilon}{2}$. Hence, for all $\epsilon>0,\left|a-b\right|\leq\left|a_{n}-a\right|+\left|a_{n}-b\right|<\epsilon$ and so $a=b$.

Now take $R$ such that $\{a_{n}:n\in\mathbb{N}\}\subseteq(-R,R)$. We will try to define by induction sequences $x_{n},y_{n},V_{n}$ such that $V_{n}=\left\{n\in\mathbb{N}:a_{n}\in\left(x_{n},y_{n}\right)\right\}\in U$. For $n=0$, we set $x_{n}=-R$, $y_{n}=+R$ and $V_{n}=\mathbb{N}$. If $V_{n}$ is defined, we can write it:

 $V_{n}=\underbrace{\{m\in\mathbb{N},a_{m}\in\left(x_{n},\frac{x_{n}+y_{n}}{2}% \right)\}}_{{V_{n}^{1}}}\cup\underbrace{\{m\in\mathbb{N},a_{m}\in\left(\frac{x% _{n}+y_{n}}{2},y_{n}\right)\}}_{{V_{n}^{2}}}\cup\underbrace{\{m\in\mathbb{N},a% _{m}=\frac{x_{n}+y_{n}}{2}\}}_{{V_{n}^{3}}}$

If $V_{n}^{3}\in U$ then we stop the construction here. Otherwise, $V_{n}=(V_{n}^{1}\cup V_{n}^{2})\cup V_{n}^{3}\in U$ so $(V_{n}^{1}\cup V_{n}^{2})\in U$ and either $V_{n}^{1}$ or $V_{n}^{2}$ is in $U$. Set $V_{{n+1}}$ to be the one which is in $U$ and set $(x_{{n+1}},y_{{n+1}}$ to be $\left(x_{n},\frac{x_{n}+y_{n}}{2}\right)$ or $\left(\frac{x_{n}+y_{n}}{2},y_{n}\right)$.

If we stopped the construction at $n$, set $a=\frac{x_{n}+y_{n}}{2}$. For all $\epsilon>0$, we have $\{m:\left|a_{m}-a\right|<\epsilon\}\supseteq V_{n}^{3}\in U$ and so $a$ is a $U$-limit. Otherwise, $x_{n}$ is nondecreasing, $y_{n}$ is nonincreasing and $\left|y_{n}-x_{n}\right|=2^{{1-n}}R\to 0$. So $x_{n}$, $y_{n}$ have a same limit $a$. For any $\epsilon>0$, consider $n$ such that $2^{{1-n}}R<\epsilon$. Then for all $m\in V_{n}$, $\left|a_{m}-a\right|\leq 2^{{1-n}}R<\epsilon$. Hence $\{m:\left|a_{m}-a\right|<\epsilon\}\supseteq V_{n}\in U$ and $a$ is a $U$-limit.

## 7.7

Let $D$ be a nonprincipal filter on $\omega$.

Suppose first $D$ is a $p$-point and $A_{0}\subseteq A_{1}\subseteq A_{2}\subseteq...$ is a decreasing sequence of elements of $D$. Define $B_{0}=\omega\setminus A_{0}$ and for all $n<\omega,B_{{n+1}}={(\omega\setminus A_{{n+1}})}\setminus{(\omega\setminus A_{% {n+1}})}.$ Then $\{B_{n}:n<\omega\}$ is a partition of $\omega$. For all $n<\omega$, $B_{n}\subseteq(\omega\setminus A_{n})\notin D$ so $B_{n}\notin D$. Hence we can find $X\in D$ such that $\forall n,X\cap B_{n}$ is finite. Then $X\setminus A_{n}=\bigcup_{{i\leq n}}(X\cap B_{i})$ is finite.

Conversely, suppose the right hand side of the equivalence holds. Let $\{B_{n}:n<\omega\}$ be a partition of $\omega$ such that $\forall n,B_{n}\notin D$. Define $A_{n}=\bigcap_{{i\leq n}}(\omega\setminus B_{i})\in D$. We have $A_{0}\subseteq A_{1}\subseteq A_{2}\subseteq...$ so there exists some $X\in D$ such that for all $n<\omega$, $X\setminus A_{n}$ is finite. So $X\cap B_{n}\subseteq X\cap\bigcup_{{i\leq n}}B_{i}=X\setminus A_{n}$ is finite too. Hence $D$ is a $p$-point.

## 7.8

Let $(P,<)$ be a countable linearly ordered set and $D$ be a $p$-point on $P$. We will consider sequences $A_{n}\in D$ and $x_{n}\in A_{n}$. We define $A_{0}=\omega$ and we choose $x_{0}$ an element of $A_{0}$. If $A_{n}$ and $x_{n}$ are defined, we have a decomposition in disjoint subsets $A_{n}=A_{n}^{-}\cup\{x_{n}\}\cup A_{n}^{+}$ where $A_{n}^{-}=\{x\in A_{n}:x and $A_{n}^{+}=\{x\in A_{n}:x>x_{n}\}$. $A_{n}\in D$ and $\{x_{n}\}\notin D$, so let $A_{{n+1}}$ be the unique subset among $A_{n}^{-},A_{n}^{+}$ which is in $D$. In particular, $A_{{n+1}}\neq\emptyset$ and we can pick $x_{{n+1}}\in A_{{n+1}}$.

By construction, we have $A_{0}\supseteq A_{1}\supseteq A_{2}\supseteq...$ where $\forall n,A_{n}\in D$. By exercise 7.7, there exists $X\in D$ such that $X\setminus A_{n}$ is finite for any $n<\omega$. We define

 $X^{\pm}=\bigcup_{{n<\omega;A_{{n+1}}=A_{n}^{\pm}}}\underbrace{(X\cap A_{n}% \setminus A_{{n+1}})}_{{X_{n}^{\pm}}}$

Each $X_{n}^{\pm}$ is finite. If $n_{0} are the integers such that $A_{{n+1}}=A_{n}^{+}$ then $X^{+}$ is order as follows: the elements of $X_{{n_{0}}}^{+}$ at the beginning, followed by the elements of $X_{{n_{1}}}^{+}$, followed by the elements of $X_{{n_{2}}}^{+}$, … So $X^{+}$ is of order type $\omega$ or a finite ordinal. Similarly, if $m_{0} are the integers such that $A_{{n+1}}=A_{n}^{-}$ then $X^{-}$ is order as follows: the elements of $X_{{n_{0}}}^{+}$ at then end, preceded by the elements of $X_{{n_{1}}}^{+}$, preceded by the elements of $X_{{n_{2}}}^{+}$, … So $X^{+}$ is of order type $\omega^{\star}$ or a finite ordinal. Clearly, $X=X^{+}\cup X^{-}$ and so one of $X^{-},X^{+}$ is in $D$ and in particular infinite. Thus there is an element of $D$ of order-type $\omega$ or $\omega^{\star}$.

## 7.9

Let $D$ be an ultrafilter.

Suppose that $D$ is Ramsey. Let $f:\omega\rightarrow\omega$. Define $A_{n}=f_{{-1}}({\{n\}})$. If one of the $A_{n}$ is in $D$, then $f$ is constant on $A_{n}\in D$. Otherwise, there exists $X\in D$ such that $\forall n,{|X\cap A_{n}|}\leq 1$ i.e. $f$ is one-to-one on $X\in D$.

Conversely, suppose the right hand side of the equivalence holds. Let $A_{n}$ be a partition of $\omega$ such that $\forall n,A_{n}\notin D$. Define $f:\omega\rightarrow\omega$ the function that associates to each $x$ the unique $n$ such that $x\in A_{n}$. $f$ is not constant on a set $X\in D$, otherwise there is $n$ such that $X\subseteq A_{n}$ and so $A_{n}\in D$ contrary to our assumption. Hence $f$ is one-to-one on a set $X\in D$ which means that $\forall n,{|X\cap A_{n}|}\leq 1$. Finally, $D$ is Ramsey.

## 7.10

Suppose $D=f_{{\star}}(D)=\{X\subseteq\omega:f_{{-1}}(X)\in D\}$.

Define $X=\{n:f(n). For each $n\in X$, the maximal decreasing sequence $n>f(n)>f(f(n))>...$ is of finite length $l(n)$. Let $X_{0}$ and $X_{1}$ be the elements of $X$ of even and odd length respectively. If $n\in f_{{-1}}(X_{0})$ then there is $m\in X_{0}$ such that $f(n)=m$. If moreover $n\in X$ then $l(n)=l(m)+1$ so $n\in X_{1}$. Hence $f_{{(-1)}}(X_{0})\cap X_{0}=\emptyset$. $X_{0}\notin D$ for otherwise by $D=f_{{\star}}(D)$ we would have $f_{{(-1)}}(X_{0})\in D$ and so $\emptyset=f_{{(-1)}}(X_{0})\cap X_{0}\in D$. Similarly, $f_{{(-1)}}(X_{1})\cap X_{1}=\emptyset$ and so $X_{1}\notin D$. Finally, $X\notin D$.

Now let $Y=\{n:f(n)>n\}$. As above, the subset $\{n\in Y:\text{any increasing sequence}\,n is not in $D$. Let’s consider the set $Z$ of elements of $Y$ such that there is an infinite increasing sequence $n. For $x,y\in Z$, define $x\equiv y$ if there are $k,m$ such that $f^{k}(x)=f^{m}(y)$. For each $x\in Z$, let $a_{x}$ be a fixed representative of the class $\{y:y\equiv x\}$ and let $l(x)$ be the least $k+m$ such that $f^{{k}}(x)=f^{{m}}(a_{x})$. Let $Z_{0}$ and $Z_{1}$ be the elements of $Z$ of even and odd length respectively. If $n\in f_{{-1}}(Z)\cap Z$ then there is $p\in Z$ such that $n. So $n\equiv p$ and $a_{p}=a_{n}$. Let $k,m,r,s$ such that $k+m=l(p),r+s=l(n),f^{{k}}(p)=f^{{m}}(a_{p}),f^{{r}}(n)=f^{{s}}(a_{n})$. If $r=0$ then $f^{{s+1}}(a_{n})=f(n)=p$ so $m\leq k+m=l(p)\leq s+1$ and so $s\geq m-1$. If $s\geq m$ we would have $n=f^{s}(a_{n})=f^{{s-m+m}}(a_{p})=f^{{s-m+k}}(p)=f^{{s-m+k+1}}(n)$ with $s-m+k+1\geq 1$ which contradicts $n\in Z$. Hence $s=m-1$ and $m=s+1\geq k+m\geq m$ implies $k=0$. We obtain $l(p)=k+m=m=s+1=l(n)+1$. Now suppose $r>0$, on the one hand we have $f^{{k+1}}(n)=f^{k}(p)=f^{{m}}(a_{n})$ i.e. $l(n)\leq k+1+m=l(p)+1$ and on the other hand, we have $f^{{r-1}}(p)=f^{{r}}(n)=f^{{s}}(a_{n})$ and so $l(p)\leq r-1+s=l(n)-1$. Hence $l(p)=l(n)-1$. Finally, $l(p)$ and $l(n)$ have different parities. So $f_{{-1}}(Z_{0})\cap Z_{0}=\emptyset$ and $f_{{-1}}(Z_{1})\cap Z_{1}=\emptyset$. As above, we deduce that $Z\notin D$ and finally $Y\notin D$.

We thus have $X\cup Y\notin D$ and $\{n:f(n)=n\}=\omega\setminus{X\cup Y}\in D$.

## 7.11

Suppose that $D\leq E$ and $E\leq D$. Then there are functions $f,g\in\omega^{\omega}$ such that $D=f_{\star}(E)=\{X\subseteq\omega:f_{{-1}}(X)\in E\}$ and $E=g_{\star}(D)=\{X\subseteq\omega:g_{{-1}}(X)\in D\}$. Hence $E=\{X\subseteq\omega:f_{{-1}}(g_{{-1}}(X))\in E\}={(g\circ f)}_{\star}(E)$. By exercise 7.10, the set $S=\{n\in\omega:g({f(n)})=n\}$ is in $E$ and in particular $S$ is infinite. $(g\circ f)_{{S}}$ is one-to-one and has image $S$ so $f_{{|S}}$ is one-to-one and $g_{{|f(S)}}$ has image $S$. Hence $\check{f}:S\rightarrow f(S)$ and $\check{g}:f(S)\rightarrow S$ are bijections (where we write $\check{f},\check{g}$ the functions with restricted domain and codomain). Let $S_{0},S_{1}$ be two disjoint infinite sets such that $S=S_{0}\cup S_{1}$ and take $T_{2}=S_{i}$ the one which is in $E$. The set $T_{1}=\check{g}_{{-1}}(T_{2})$ is in $D$ because $f_{{-1}}(T_{1})\supseteq f_{{-1}}(g_{{-1}}(T_{2}))\in E$. Moreover $\omega\setminus T_{2}\supseteq S_{{1-i}}$ and $\omega\setminus T_{1}\supseteq\check{g}_{{-1}}(S_{{1-i}})$ are infinite and thus there is a bijection between $\omega\setminus T_{1}$ and $\omega\setminus T_{2}$. Thus the bijection between $T_{1}$ and $T_{2}$ provided by $\check{g}$ can be extended to a bijection $h\in\omega^{\omega}$.

Now, if $X\in E$ then $X\cap T_{2}\in E$. Thus $h_{{-1}}(X\cap T_{2})=\check{g}_{{-1}}(X\cap T_{2})=g_{{-1}}(X\cap T_{2})\in D$. But $h_{{-1}}(X)\supseteq h_{{-1}}(X\cap T_{2})$, and so $h_{{-1}}(X)\in D$. Then $g_{{-1}}(X)\cap D=\check{g}_{{-1}}(X)\cap D=h_{{-1}}(X)\cap D\in D$. Because $g_{{-1}}(X)\supseteq g_{{-1}}(X)\cap D$ we get $g_{{-1}}(X)\in D$. Finally, our assumption $E=g_{\star}(D)$ allows to come back to $X\in E$. With this sequence of implications, we have proved that $E=\{X\subseteq\omega:h_{{-1}}(X)\in D\}=h_{\star}(D)$. Since $h:\omega\rightarrow\omega$ is a bijection, this means $D\equiv E$.

## 7.12

(We use the characterisation of Ramsey ultrafilters given in exercise 7.9)

Suppose $D$ is a Ramsey ultrafilter. And suppose $E$ is a nonprincipal ultrafilter such that $E\leq D$. There is $f:\omega\rightarrow\omega$ such that $E=f_{\star}(D)=\{X\subseteq\omega:f_{{-1}}(X)\in D\}$ There exists $X\in D$ such that $f$ is either constant or one-to-one on $X$. $f$ is not constant: if $f(\omega)$ is a singleton then $Y=\omega\setminus f(\omega)\in E$ (because $E$ is nonprincipal) and so $\emptyset=f_{{-1}}(Y)\in D$. Thus $f$ is one-to-one on $X$ and gives a bijection from $X$ to $f(X)$ that can be extended to a bijection $h\in\omega^{\omega}$. But $f_{{-1}}(f(X))\subseteq X$ and so $f(X)\in D$. As in 7.11, we get $E=h_{\star}(D)$ and so $D\equiv E$.

Now suppose $D$ is a minimal ultrafilter on $\omega$. Let $f:\omega\rightarrow\omega$ such that for any $X\in D$ the restriction $f_{{|X}}$ is not constant. Let $Y_{0}\in f_{{\star}}(D)$ and $X_{0}\in D$ such that $X_{0}=f_{{-1}}(Y_{0})$. Let $m\in X_{0}$ (and so $f(m)\in Y_{0}$) and $X_{1}=X_{0}\cap f_{{-1}}(\{{f(m)}\})$. $m\in X_{1}\neq\emptyset$ and because $f_{{|X_{1}}}$ is contant we have $X_{1}\notin D$. Then $f_{{-1}}(Y_{0}\setminus\{f(m)\})=X_{0}\setminus X_{1}\in D$ and consequently $Y_{0}\setminus\{f(m)\}\in D$. Hence $f_{{\star}}(D)$ is nonprincipal. By definition, $f_{\star}(D)\leq D$ and by minimality $f_{\star}(D)\equiv D$. Hence there is a bijection $g$ such that $D=g_{\star}(f_{\star}(D))$. Using exercise $D$ we find that $X=\{n:g(f(n))=n\}\in D$. $(g\circ f)_{{|X}}$ is a bijection and so $f_{{|X}}$ is one-to-one on $X\in D$.

## 7.13

Let $\omega_{\alpha}$ be a regular cardinal and suppose $I$ is a nonprincipal $\omega_{\alpha}$-complete ideal on $\omega_{\alpha}$. Let $I=\{X_{\delta}:\delta<\lambda$ be an enumeration of $I$. Because $I$ is nonprincipal, we have $\bigcup_{{\delta<\lambda}}=\omega_{\alpha}$. Let $\{Y_{\gamma}\subseteq\omega_{\alpha}:\gamma<\operatorname{cf}(\omega_{\alpha})\}$ such that $\bigcup_{{\gamma<\operatorname{cf}(\omega_{\alpha})}}Y_{\gamma}=\omega_{\alpha}$. Define for all $\gamma<\operatorname{cf}(\omega_{\alpha})$ the ordinal $\delta_{\gamma}$ to be least $\delta$ such that $\bigcup_{{\beta<\delta}}Y_{\beta}\subseteq\bigcup_{{\beta<\delta}}X_{\beta}$. By construction, we have $\omega_{\alpha}=\bigcup_{{\gamma<\operatorname{cf}(\omega_{\alpha})}}\bigcup_{% {\beta<\delta_{\gamma}}}X_{\beta}$. But each $X_{\beta}\in I$, $\delta_{\gamma}<\operatorname{cf}(\omega_{\alpha})<\omega_{\alpha}$ and $I$ is $\omega_{\alpha}$-complete: hence this union is in $I$. A contradiction.

## 7.14

Let $I$ be the set of borel sets of Lebesgue measure 0. We have $\mu(\emptyset)=0$, $\mu(\mathbb{R})=\infty$. If $X\subseteq Y$ are borel sets and $\mu(Y)=0$ then $0\leq\mu(X)\leq\mu(Y)=0$. Finally, if $\{X_{n}:n\in\mathbb{N}\}$ are borel sets, then so are $Y_{n}=X_{n}\setminus\bigcup_{{i and $0\leq=\mu\left({\bigcup_{{n\in\mathbb{N}}}X_{n}}\right)=\mu\left({\bigcup_{{n% \in\mathbb{N}}}Y_{n}}\right)=\sum_{{n\in\mathbb{N}}}\mu(Y_{n})\leq\sum_{{n\in% \mathbb{N}}}\mu(X_{n})=0$. Hence $I$ is a $\sigma$-ideal.

(We have followed the statement given page 82. Should the proof be generalized to non-borel sets of measure 0?)

## 7.15

We have $\emptyset=\bigcup_{{n\in\mathbb{N}}}\emptyset$ and $\mathbb{R}$ is no meager by the Baire Category Theorem. If we $X_{n}$ is a countable family of meager sets then the union is still a countable union of nowhere dense sets and so is meager. Finally, if $X\subseteq Y$, $X=\bigcup_{{n\in\mathbb{N}}}X_{n}$ and $Y=\bigcup_{{n\in\mathbb{N}}}Y_{n}$ where each $X_{n},Y_{n}$ is meager, we have $X=X\cap Y=\bigcup_{{n,m\in\mathbb{N}}}X_{n}\cap Y_{m}$ and each $X_{n}\cap Y_{m}$ is nowhere dense (e.g included in the nowhere dense set $X_{n}$).

## 7.16

As in 7.2, we prove that $F$ is a nonprincipal filter on $S$. Moreover, suppose $\lambda<\kappa$ and we have a sequence $\{X_{\alpha}\in F:\alpha<\lambda\}$. Let $P_{\alpha}\in S$ such that $\forall\alpha<\lambda,X_{\alpha}\supseteq\hat{P_{\alpha}}$. Then

 $\bigcap_{{\alpha<\lambda}}X_{\alpha}\supseteq\bigcap_{{\alpha<\lambda}}\hat{P_% {\alpha}}=\hat{\bigcup_{{\alpha<\lambda}}P_{\alpha}}$

And because $\kappa$ is regular, $\bigcup_{{\alpha<\lambda}}P_{\alpha}\in S$. Hence $F$ is $\kappa$-complete.

## 7.17

Most of the properties are obvious. Let’s prove the associativity of $\oplus$ and the distributivity of $\cdot$ over $\oplus$. We have $u\oplus(v\oplus w)=u\cdot{-(v\cdot-w+-v\cdot w)}+-u\cdot{(v\cdot-w+-v\cdot w)}% =u\cdot{\left((-v+w)\cdot(v\cdot-w)\right)}-u\cdot{(v\cdot-w+-v\cdot w)}=u% \cdot-v\cdot-w+u\cdot w\cdot v+-u\cdot v\cdot-w+-u\cdot-v\cdot w$. This expression is symmetric in $u,v,w$ and $\oplus$ is clearly commutative so $\oplus$ is associative. Similarly, $u\cdot(v\oplus w)=u\cdot{(v\cdot-w+-v\cdot w)}=u\cdot v\cdot-w+u\cdot-v\cdot w% =u\cdot v\cdot{(-u+-w)}+(-u+-v)\cdot u\cdot w=u\cdot v\cdot{-(u\cdot w)}+-{(u% \cdot-v)}\cdot u\cdot w=(u\cdot v)\oplus(u\cdot w)$

## 7.18

Let $X\subseteq B$ and $A$ the subalgebra generated by $X$. Let $Y$ be the set of boolean combinations as defined in the exercise. We have $X\subseteq A$ and $A$ is stable by boolean operations so $Y\subseteq A$.

For the inverse inclusion, it suffices not note that $Y$ is a subalgebra. $0,1\in Y$ as empty sum and product. If we consider boolean operations on elements of $Y$, then it is clear that using the usal properties of the boolean algebra (absorption, distributivity…) we can put the result in the form of elements of $Y$.

If $X$ is infinite, first it is clear $|X|\leq|A|$. The inverse equality is obtained by considering that

 $|A|\leq\sum_{{n<\omega}}{\left(\sum_{{m<\omega}}{{(2|X|)}^{n}}\right)}^{m}=|X|$

## 7.19

Let $A\subseteq B$, $u\in B$ and define $C={a.u+b.(-u):a,b\in A}$ and $D$ the subalgebra generated by $A\cup\{u\}$. We shall prove that $C=D$.

First, we notice that for $a=1$ and $b=0$, we get $u\in C$. If we consider $a=b$, we get $A\subseteq C$. So to show $D\subseteq C$ it is enough to prove that $C$ is a subalgebra. We already have $0,1\in A\subseteq C$. Moreover, we see that $C$ is stable by boolean operations, if $a_{1},a_{2},b_{1},b_{2}\in A$ we have:

 $\displaystyle{\left(a_{1}.u+b_{1}.(-u)\right)}+{\left(a_{2}.u+b_{2}.(-u)\right)}$ $\displaystyle={(a_{1}+a_{2}).u+(b_{1}+b_{2}).(-u)}$ $\displaystyle{\left(a_{1}.u+b_{1}.(-u)\right)}.{\left(a_{2}.u+b_{2}.(-u)\right)}$ $\displaystyle={(a_{1}.a_{2}).u+(b_{1}.b_{2}).(-u)}$ $\displaystyle-{\left(a_{1}.u+b_{1}.(-u)\right)}$ $\displaystyle={\left(-a_{1}+(-u)\right)}.{\left(-b_{1}+u\right)}$ $\displaystyle={(-a_{1}).(-b_{1})}+{(-a_{1}).u}+{(-b_{1}).(-u)}$

## 7.20

By induction on $k=|X|$. If $k=0$ then $A={0,1}$ and $|A|=2\leq 2^{{2^{0}}}$. If the result is true for some $k\geq 0$ and we add an element $u$ to $A$ then the boolean algebra generate by $A\cup\{u\}$ is of size at most $2^{{2^{k}}}2^{{2^{k}}}=2^{{2^{k}+2^{k}}}=2^{{22^{k}}}=2^{{2^{{k+1}}}}$.

## 7.21

Let $B$ be a finite boolean algebra. For all $x\in B$, any sequence $x=u_{0}>u_{1}>u_{2}>...$ is finite, so $x$ is atomic. Let $A=\{a_{1},a_{2},...,a_{n}\}$ be the set of atoms of $B$. Then we have for all $1\leq i\neq j\leq n$, $a_{i}\cdot a_{j}\in\{0,a_{i}\}\cap\{0,a_{j}\}=\{0\}$ i.e. $a_{i}\cdot a_{j}=0$. Consequently, for all $1\leq j\leq n$ we have $a_{j}\cdot\sum_{{i=1}}^{n}a_{i}=\sum_{{i=1}}^{n}(a_{j}\cdot a_{i})=a_{j}\neq 0$ and so $a_{j}\nleq-{\sum_{{i=1}}^{n}a_{i}}$ and $-\sum_{{i=1}}^{n}a_{i}=0$ i.e. $\sum_{{i=1}}^{n}a_{i}=1$.

Define the application $\phi:\operatorname{\mathcal{P}}(A)\rightarrow B$ by $\phi(X)=\sum_{{i=1}}^{n}{\chi_{X}(a_{i})}a_{i}$. We have $\phi(\emptyset)=0$ and $\phi(A)=\sum_{{i=1}}^{n}a_{i}=1$. Let $X,Y\subseteq A$. We have

 $\displaystyle\phi(X)\cdot\phi(Y)$ $\displaystyle=\left(\sum_{{i=1}}^{n}\chi_{{X}}(a_{i})a_{i}\right)\left(\sum_{{% i=1}}^{n}\chi_{{Y}}(a_{i})a_{i}\right)$ $\displaystyle=\sum_{{i=1}}^{n}\chi_{{X}}(a_{i})\chi_{{Y}}(a_{i})a_{i}$ $\displaystyle=\sum_{{i=1}}^{n}\chi_{{(X\cap Y)}}(a_{i})a_{i}$ $\displaystyle=\phi(X\cap Y)$

The sum $\chi_{X}(a_{i})+\chi_{Y}(a_{i})$ takes the values $a_{i}+a_{i}=a_{i},a_{i}+0=a_{i},0+0=0$ according to whether $a_{i}$ is in $X\cap Y$, $X\triangle Y$ or outside $X\cup Y$. Hence

 $\displaystyle\phi(X)+\phi(Y)$ $\displaystyle=\sum_{{i=1}}^{n}(\chi_{X}(a_{i})+\chi_{Y}(a_{i}))a_{i}$ $\displaystyle=\sum_{{i=1}}^{n}\chi_{{(X\cup Y)}}(a_{i})a_{i}$ $\displaystyle=\phi(X\cup Y)$

Finally, we note that $\phi(A\setminus X)=-\phi(X)$ because $\phi(A\setminus X)+\phi(X)=\sum_{{i=1}}^{n}{(\chi_{{(A\setminus X)}}(a_{i})+% \chi_{{X}}(a_{i}))a_{i}}=\sum_{{i=1}}^{n}a_{i}=1$ and $\phi(A\setminus X)\cdot\phi(X)=\sum_{{i=1}}^{n}{(\chi_{{(A\setminus X)}}(a_{i}% )\chi_{{X}}(a_{i}))a_{i}}=0$. Hence $\phi$ is a homomorphism of boolean algebras.

Suppose $\phi(X)=\phi(Y)$. Then $\sum_{{i=1}}^{n}\chi_{{X}}(a_{i})a_{i}=\sum_{{i=1}}^{n}\chi_{{Y}}(a_{i})a_{i}$. So for all $1\leq j\leq n$ then $\chi_{{X}}(a_{j})a_{j}=a_{j}\cdot\sum_{{i=1}}^{n}\chi_{{X}}(a_{i})a_{i}=a_{j}% \cdot\sum_{{i=1}}^{n}\chi_{{Y}}(a_{i})a_{i}=\chi_{{Y}}(a_{j})a_{j}$ and thus $\chi_{{X}}(a_{i})=\chi_{{Y}}(a_{i})$. Hence $X=Y$ and $\phi$ is one-to-one.

Let $b\in B$ and $X=\{n:a_{n}\leq b\}$. Then $\forall n\in X,a_{n}\leq b$ so $\phi(X)=\sum_{{n\in X}}a_{n}\leq b$. If $\phi(X) then $b-\phi(X)\neq 0$. Thus there exists an atom $a_{j}$ such that $a_{j}\leq b-\phi(X)$ i.e. $a_{j}\leq b$ and $a_{j}\leq-\phi(X)=\sum_{{i=1}}^{n}\chi_{{(A\setminus X)}}(a_{i})a_{i}=\sum_{{i% =1}}^{n}{(1-\chi_{X}(a_{i}))}a_{i}$. Then $a_{j}=a_{j}\cdot a_{j}\leq{(1-\chi_{X}(a_{j}))}a_{j}$ and so $1-\chi_{X}(a_{j})=1$ i.e. $\chi_{X}(a_{j})=0$ and $a_{j}\notin X$. This contradicts $a_{j}\leq b$. So $\phi(X)=b$ and $\phi$ is surjective.

Finally, $\phi:\operatorname{\mathcal{P}}(A)\cong B$ is an isomorphism of algebras. In particular $|B|=|\operatorname{\mathcal{P}}(A)|=2^{n}$.

## 7.23

Let $\phi:B\rightarrow B_{{|a}}$ such that $\phi(x)=x\cdot a$. We have $\phi(x\cdot y)=x\cdot y\cdot a=x\cdot a\cdot y\cdot a=\phi(x)\cdot\phi(y)$, $\phi(x+y)=(x+y)\cdot a=x\cdot a+y\cdot a=\phi(x)+\phi(y)$, $\phi(0)=0\cdot a=0$, $\phi(1)=1\cdot a=a$ and $\phi(-x)=-x\cdot a=a-x$. Hence $\phi$ is a homomorphism of boolean algebras. For all $x\in B_{{|a}}$, $x\leq a$ and so $\phi(x)=x\cdot a=x$. Hence $\phi(B)=B_{{|a}}$. Moreover $\phi(x)=0\Leftrightarrow x\cdot a=0\Leftrightarrow x\leq-a\Leftrightarrow x\in I% =\{u,u\leq-a\}$. Hence $\ker\phi=I$ and $B_{{|a}}\cong B/I$.

## 7.24

Let $A$ be a subalgebra of $B$ and $u\in B\setminus A$. Define $H=\{a\in A:u. For all $a,b\in H$ we have either $a\cdot b=u\neq 0$, $a\cdot b=-u\neq 0$ or e.g. $u and $-u. In that latter case, we have $a\nleq u$ and so $a\cdot b\geq a\cdot-u\neq 0$. Hence $H$ has the finite intersection property and generates a filter on $A$ that can be extended to an ultrafilter $\hat{H}\subseteq A$. Note that if $a\in\hat{H}$ then $u\cdot a\neq 0$ or otherwise $u\leq-a$ and so $-a\in\hat{H}$. Consequently, $\hat{H}\cup\{u\}$ has the finite intersection property and generates a filter on that can be extended to an ultrafilter $F$ on $B$. Similarly, we get an ultrafilter $G\supseteq\hat{H}\cup\{-u\}$ on $B$. Then by definition, $u\in F$ and $u\notin G$. Moreover, because $\hat{H}$ is an ultrafilter on $A$ we have $a\in\hat{H}$ or $-a\in\hat{H}$ for all $a\in A$ and so $F\cap A=G\cap A=\hat{H}$.

## 7.25

Let $B$ be an infinite boolean algebra and $S$ the set of ultrafilter on $B$. Suppose $|S|<|B|$. For all $F,G\in S$ such that $F\neq G$ we have $F\nsubseteq G$ because $F$ is maximal and so $F\setminus G\neq 0$. For each such pairs $F\neq G$, pick $u\in F\setminus G$ to form a set $X$ and let $A$ be the subalgebra generated by $X$. Hence for all $F\neq G$, we have some $u\in F\cap X\subseteq F\cap A$ such that $u\notin G\supseteq G\cap A$ and so $F\cap A\neq G\cap A$.

If $A$ is infinite then by exercise 7.18, $|A|=|X|\leq{|S|}^{2}<|B|$ and if $A$ is finite, $|A|<|B|$ is obvious. So there is $u\in B\setminus A$. By exercise 7.24, we can use this element $u$ to construct ultrafilters $F\neq G$ such that $F\cap A=G\cap A$. Contradiction.

## 7.26

Let $B$ a boolean algebra such that $\sum$ is defined for any subset. Let $X\subseteq B$ and define $Y=\{y\in B:\forall x\in X,y\leq x\}$. If $x\in X$ then $\forall x\in X,y\leq x$ and so $\sum Y\leq x$. If $z$ is a lower bound of $X$ i.e. $\forall x\in X,z\leq x$ then $z\in Y$ and so $z\leq\sum Y$. Finally $\sum Y$ is the greatest lower bound of $X$ and $\prod X$ is defined for all $X$.

## 7.27

Let $B$ a complete boolean algebra $X\subseteq B$ and $a\in B$.

1. (1)

For all $u\in X$ we have $u\leq{\sum X}$ and so $a\cdot u\leq a\cdot{\sum X}$. Thus $\sum{a\cdot X}\leq a\cdot{\sum X}$. For all $u\in X$ we have $a\cdot u\leq u\leq{\sum X}$ so ${\sum a\cdot X}\leq{\sum X}$. Moreover, $a\cdot u\leq a$ so ${\sum a\cdot X}\leq a$. Hence ${\sum a\cdot X}\leq a\cdot{\sum X}$.

2. (2)

For all $u\in X$ we have $u\geq{\prod X}$ and so $a+u\geq a+{\prod X}$. Thus $\prod{a+X}\geq a+{\prod X}$. For all $u\in X$ we have $a+u\geq u\geq{\prod X}$ so ${\prod a+X}\geq{\prod X}$. Moreover, $a+u\geq a$ so ${\prod a+X}\geq a$. Hence ${\prod a+X}\geq a+{\prod X}$.

3. (3)

For all $u\in X$ we have $u\leq{\sum X}$ and so $-{\sum X}\leq-u$ and $-{\sum X}\leq{\prod-X}$. Moreover, ${\prod-X}\leq-u$, $u\leq-{\prod-X}$, ${\sum X}\leq-{\prod-X}$. Finally, $-{\sum X}={\prod-X}$.

4. (4)

For all $u\in X$ we have $u\geq{\prod X}$ and so $-{\prod X}\geq-u$ and $-{\prod X}\geq{\sum-X}$. Moreover, ${\sum-X}\geq-u$, $u\geq-{\sum-X}$, ${\prod X}\geq-{\sum-X}$. Finally, $-{\prod X}={\sum-X}$.

## 7.29

Let $A$ be a subalgebra of a Boolean algebra $B$. Let $u\in B$ and $A(u)$ be the algebra generated by $A\cup\{u\}$. Let $h$ be a homomorphism from $A$ into a complete Boolean algebra $C$.

Let $a,b\in A$ such that $a\leq u\leq b$ then $h(a)\leq h(b)$. Thus for all $b\in A$ such that $u\leq b$, we have $\sum\{h(a):a\in A,a\leq u\}\leq h(b)$ and finally $\sum\{h(a):a\in A,a\leq u\}\leq\prod\{h(b):b\in A,u\leq b\}$. Let’s choose $v\in C$ is such that $\sum\{h(a):a\in A,a\leq u\}\leq v\leq\prod\{h(b):b\in A,u\leq b\}$.

Let $a_{1},a_{2},b_{1},b_{2}\in A$ such that $a_{1}\cdot u+b_{1}\cdot-u=a_{2}\cdot u+b_{2}\cdot-u$. If we multiply both sides by $-a_{2}\cdot u$ and $-b_{2}\cdot-u$ we get two new equalities $-a_{2}\cdot a_{1}\cdot u=0$ and $-b_{2}\cdot b_{1}\cdot-u=0$ respectively. This means $-a_{2}\cdot a_{1}\leq-u$ and $-b_{2}\cdot b_{1}\leq u$ respectively. Now from $-b_{2}\cdot b_{1}\leq u\leq a_{2}+-a_{1}$ and the definition of $v$, we get $-h(b_{2})\cdot h(b_{1})\leq v\leq h(a_{2})+-h(a_{1})$. Hence $v\leq h(a_{2})+-h(a_{1})$ and $-v\leq h(b_{2})+-h(b_{1})$. If we multiply the first inequality by $h(a_{1})\cdot v$ and the second by $h(b_{1})\cdot-v$ we get $h(a_{1})\cdot v\leq h(a_{1})\cdot h(a_{2})\cdot v$ and $h(b_{1})\cdot-v\leq h(b_{1})\cdot h(b_{2})\cdot-v$ respectively. A fortiori, $h(a_{1})\cdot v\leq h(a_{2})\cdot v$ and $h(b_{1})\cdot-v\leq h(b_{2})\cdot-v$. By symmetry, we have $h(a_{1})\cdot v=h(a_{2})\cdot v$, $h(b_{1})\cdot-v=h(b_{2})\cdot-v$ and finally $h(a_{1})\cdot v+h(b_{1})\cdot-v=h(a_{2})\cdot v+h(b_{2})\cdot-v$.

By the previous analysis, for any $x\in A(u)$ we can define $f(x)=h(a)\cdot v+h(b)\cdot-v$ if $x=a\cdot u+b\cdot-u$ and this does not depend on the decomposition of $x$. In particular for all $a\in A$, we have $f(a)=f(a\cdot u+a\cdot-u)=h(a)\cdot v+h(a)\cdot-v=h(a)$ and so $f(0)=h(0)=0$ and $f(1)=h(1)=1$.

 $\displaystyle f((a_{1}\cdot u+b_{1}\cdot-u)+(a_{2}\cdot u+b_{2}\cdot-u))$ $\displaystyle=f((a_{1}+a_{2})\cdot u+(b_{1}+b_{2})\cdot-u)$ $\displaystyle=h(a_{1}+a_{2})\cdot v+h(b_{1}+b_{2})\cdot-v$ $\displaystyle=(h(a_{1})\cdot v+h(b_{1})\cdot-v)+(h(a_{2})\cdot v+h(b_{2})\cdot% -v)$ $\displaystyle=f(a_{1}\cdot u+b_{1}\cdot-u)+f(a_{2}\cdot u+b_{2}\cdot-u)$
 $\displaystyle f((a_{1}\cdot u+b_{1}\cdot-u)\cdot(a_{2}\cdot u+b_{2}\cdot-u))$ $\displaystyle=f(a_{1}\cdot a_{2}\cdot u+b_{1}\cdot b_{2}\cdot-u)$ $\displaystyle=h(a_{1}\cdot a_{2})\cdot v+h(b_{1}\cdot b_{2})\cdot-v$ $\displaystyle=(h(a_{1})\cdot v+h(b_{1})\cdot-v)\cdot(h(a_{2})\cdot v+h(b_{2})% \cdot-v)$ $\displaystyle=f(a_{1}\cdot u+b_{1}\cdot-u)\cdot f(a_{2}\cdot u+b_{2}\cdot-u)$

Finally we have

 $\displaystyle f(-(a\cdot u+b\cdot-u))$ $\displaystyle=f((-a+-u)\cdot(-b+u))$ $\displaystyle=f((-a\cdot-b+-a)\cdot u+(-a\cdot-b+-b)\cdot-u)$ $\displaystyle=f(-a\cdot u+-b\cdot-u)$ $\displaystyle=-h(a)v+-h(b)\cdot-v$ $\displaystyle f(-(a\cdot u+b\cdot-u))+f(a\cdot u+b\cdot-u)$ $\displaystyle=(-h(a)+h(a))v+(-h(b)+h(b))\cdot-v$ $\displaystyle=v+-v$ $\displaystyle=1$ $\displaystyle f(-(a\cdot u+b\cdot-u))\cdot f(a\cdot u+b\cdot-u)$ $\displaystyle=(-h(a)\cdot h(a))v+(-h(b)\cdot h(b))\cdot-v$ $\displaystyle=0$

and so $f(-(a\cdot u+b\cdot-u))=-f(a\cdot u+b\cdot-u)$. Finally, $f$ is a homomorphism from $A(u)$ to $C$ that extends $h$.

## 7.30

Let $A$ be a subalgebra of a Boolean algebra $B$. Let $h$ be a homomorphism from $A$ into a complete Boolean algebra $C$. Let $P$ be the set of all Boolean algebra homomorphisms from a subalgebra of $B$ to $C$ which extends $h$. $(P,\subseteq)$ is a nonempty partially ordered set. If $C$ is a chain and $f_{c}=\bigcup C$ then $B_{C}=\operatorname{dom}f_{C}=\bigcup_{{f\in C}}\operatorname{dom}f$ is a subalgebra of $B$. We have $B_{C}\supseteq A$ and ${f_{c}}_{{|A}}=h$ so $f_{c}\in P$. Moreover $\forall f\in C,f_{c}\supseteq f$. By Zorn’s Lemma $P$ has a maximal element $f_{{\text{m}ax}}$. If $\operatorname{dom}f_{{\text{m}ax}}\subsetneq B$ then let $u\in B\setminus{\operatorname{dom}f}$. Using exercise 7.29 we can construct a Boolean homomorphism $g:A(u)\rightarrow C$ and we would have $g\supsetneq f_{{\text{m}ax}}$. So $f_{{\text{m}ax}}:B\rightarrow C$ is a Boolean homomorphism from $B$ into $C$ such that ${f_{{\text{m}ax}}}_{{|A}}=h$.

## 7.31

Let $B$ be a Boolean algebra and $A$ be a regular subalgebra of $B$. Denote by $i_{A}:A\rightarrow\tilde{A}$ and $i_{B}:B\rightarrow\tilde{B}$ the one-to-one mapping from $A,B$ to their respective completions. By exercise 7.30, ${i_{B}}_{{|A}}:A\rightarrow\tilde{B}$ can be extended to a mapping $i_{{\tilde{A}}}:\tilde{A}\rightarrow\tilde{B}$ such that ${i_{B}}_{{|A}}=i_{{\tilde{A}}}\circ i_{A}$.

Let $x,y\in\tilde{A}$ and suppose $x\neq y$. Then we have either $x\nleq y$ or $y\nleq x$. If for example the first assertion holds, then $0 and because $i_{{A}}(A)$ is a dense subalgebra, there exists $a\in A$ such that $0. But $a\neq 0$ because $i_{A}$ is one-to-one and so $i_{B}(a)\neq 0$ because $i_{B}$ is one-to-one. Hence $0. Thus $i_{{\tilde{A}}}(x)\nleq i_{{\tilde{A}}}(y)$. Finally $i_{{\tilde{A}}}\neq i_{{\tilde{A}}}(y)$ and $i_{{\tilde{A}}}$ is an embedding.

Using the density of $i_{{\tilde{A}}}(A)$ in $\tilde{A}$, we can construct for each $a\in\tilde{A}$ a partition $V_{a}\subseteq A$ of $a=\sum i_{A}(V_{a})$ by ordinal induction as follows. For any ordinal $\alpha$, if $v_{{\beta}}$ is defined for all $\beta<\alpha$ and if $a-\sum_{{\beta<\alpha}}i_{A}(v_{\beta})>0$ then let $v_{{\alpha}}\in A$ such that $i_{A}(v_{{\alpha}})\leq a-\sum_{{\beta<\alpha}}i_{A}(v_{\beta})$. By construction, the $v_{\alpha}$ are disjoint. By Replacement, there is $\alpha$ such that $a-\sum_{{\beta<\alpha}}i_{A}(v_{\beta})=0$ and we define $V_{a}=\{v_{\beta}:\beta<\alpha\}$. Note that $V_{0}=\emptyset$.

Now let $X=\{x_{\alpha}:\alpha<\lambda\}\subseteq A$ and denote $\sigma=\sum^{{\tilde{A}}}X$. For all $\alpha<\lambda$, let $y_{\alpha}=x_{\alpha}-\sum_{{\beta<\alpha}}y_{\beta}$ so that the $y_{\alpha}$ are pairwise disjoint and $\sigma=\sum_{{\alpha<\lambda}}y_{\alpha}$. Let $V=\bigcup_{{\alpha<\beta}}V_{{y_{\alpha}}}$ and $W=V_{{-\sigma}}$. Clearly, $\sum V=\sigma$, $i_{A}(V\cdot W)=i_{A}(V)\cdot i_{A}(W)\leq\sigma\cdot-\sigma=0$ and $\sum^{{\tilde{A}}}{i_{A}(V\cup W)}={\sum^{{\tilde{A}}}i_{A}(V)}+{\sum^{{\tilde% {A}}}i_{A}(W)}=\sigma+-\sigma=1$. Hence $1=\sum^{{\tilde{A}}}i_{A}(V\cup W)\leq\sum^{{A}}(V\cup W)$ and $V\cup W$ is a maximal antichain in $A$. Because $A$ is a regular subalgebra of $B$ and $i_{{B}}(B)$ is a regular subalgebra of $\tilde{B}$, we have $1=\sum^{{A}}(V\cup W)=\sum^{{B}}(V\cup W)=\sum^{{\tilde{B}}}i_{{B}}(V\cup W)$.

We have $\forall x\in X,x\leq\sum X$ so $\forall x\in X,i_{{\tilde{A}}}(x)\leq i_{{\tilde{A}}}(\sum X)$ and finally $\sum{i_{{\tilde{A}}}(X)}\leq i_{{\tilde{A}}}(\sum X)$. Similarly, any $v\in V$ belongs to some $V_{\alpha}$ and by construction $i_{A}(v)\leq y_{\alpha}\leq x_{\alpha}\in X$ so $i_{{\tilde{A}}}(i_{A}(v))\leq i_{{\tilde{A}}}(x)\leq{\sum i_{{\tilde{A}}}(X)}$ and we get $\sum i_{{\tilde{A}}}(i_{A}(V))\leq{\sum i_{{\tilde{A}}}(X)}$. Thus we have

 $\displaystyle i_{{\tilde{A}}}\left(\sum X\right)$ $\displaystyle={i_{{\tilde{A}}}(\sigma)}\cdot{\sum i_{{B}}(V\cup W)}$ $\displaystyle=\sum_{{w\in V\cup W}}{i_{{\tilde{A}}}(\sigma)\cdot i_{{\tilde{A}% }}(i_{A}(w))}$ $\displaystyle=\sum_{{w\in V\cup W}}{i_{{\tilde{A}}}\left(\sum_{{v\in V}}(i_{A}% (v)\cdot i_{A}(w))\right)}$ $\displaystyle=\sum_{{w\in V\cup W}}{i_{{\tilde{A}}}\left(\sum_{{v\in V}}\delta% _{{v,w}}i_{A}(w)\right)}$ $\displaystyle=\sum i_{{\tilde{A}}}(i_{A}(V))$ $\displaystyle\leq\sum i_{{\tilde{A}}}(X)$

Finally $i_{{\tilde{A}}}\left(\sum X\right)=\sum i_{{\tilde{A}}}(X)$ and $i_{{\tilde{A}}}$ is a complete embedding from $\tilde{A}$ to $\tilde{B}$ such that ${i_{B}}_{{|A}}=i_{{\tilde{A}}}\circ i_{A}$. Note that for any $a\in\tilde{A}$, such a morphism $f$ must satisfy $f(a)=f\left(\sum i_{A}(V_{a})\right)=\sum f(i_{A}(V_{a}))=\sum i_{B}(V_{a})$ and so $i_{{\tilde{A}}}$ is the unique morphism with this property.

## 7.32

Let $B$ be a complete Boolean algebra. If $W$ is a maximal antichain then any element $b\in B$ can be written $b=b\cdot 1=b\cdot\sum W=\sum_{{w\in W}}b_{w}$ where $b_{w}=b\cdot w\in B_{{w}}$. Conversely, if $\sum_{{w\in W}}b_{w}=\sum_{{w\in W}}c_{w}$ with $b_{w},c_{w}\in B_{{w}}$ then $b_{w}=w\cdot\sum_{{w\in W}}b_{w}=w\cdot\sum_{{w\in W}}c_{w}=c_{w}$ and the decomposition is unique. Hence $|B|=\prod_{{w\in W}}|B_{{|w}}|$.

The case $B$ finite has been studied in exercise 7.21. $W$ was the set of the $n$ atoms. For any $w\in W$, $B_{{|w}}=\{0,w\}$ and $|B|=\prod_{{w\in W}}|B_{{|w}}|=2^{n}$. Below we consider the case $B$ infinite and prove that ${|B|}^{{\aleph_{0}}}=|B|$.

First consider the case when $|B_{{a}}|=|B|$ for all $a\neq 0$. $B$ is infinite so there exists some $0. If $u_{n}$ is defined, then $B_{{|-u_{n}}}$ is infinite and there exists some $0… If $i then $u_{i}\cdot u_{j}\leq u_{i}\cdot-u_{i}=0$. If we set $u_{0}=-\sum_{{n\in\mathbb{N}}}u_{n}$ then $W=\{u_{n}:n\in\mathbb{N}\}$ is a maximal antichain. Hence $|B|=\prod_{{w\in W}}{|B_{{|w}}|}={|B|}^{{\aleph_{0}}}$.

Let $S$ be the set of stable element i.e. for all $0, $|{B_{{|x}}}|=|{B_{{|a}}}|$. Let $b\in B$. Suppose that there is no $0 such that $B_{{|x}}$ is stable then there is $0 such that $|{B_{{|u_{1}}}}|<|{B_{{|u_{0}}}}|$. Thus we could continue and construct an infinite decreasing sequence of cardinals $|B_{{|u_{0}}}|>|B_{{|u_{1}}}|>|B_{{|u_{2}}}|>...$. A contradiction. So $\exists x\in S,x\leq b$ and $S$ is dense in $B$. Using the technique of the exercise 7.31, we can construct a maximal antichain $W\subseteq S$ such that $|W|\leq\aleph_{0}$.

Let $s\in S$. If $s$ is an atom then $B_{{|s}}=\{0,s\}$. Otherwise, there exists some $0. We have $|B_{{|s}}|=|B_{{|x}}|$ and $B_{{|x}}\subsetneq B_{{|s}}$ so $B_{{|s}}$ is infinite. By the first case, $|B_{{|s}}|={|B_{{|x}}|}^{{\aleph_{0}}}$.

If $\forall w\in W$, $|B_{{|w}}|=2$ then $|B|=2^{{|W|}}$ and so $W$ is infinite: $|W|=\aleph_{0}$. Clearly, we have ${|B|}^{{\aleph_{0}}}=|B|$. Otherwise, there is some $w_{0}$ such that $|B_{{|w_{0}}}|$ is infinite. Let $T\subseteq W$ the set of atoms of $B$. We have $|T|\leq|W|\leq\aleph_{0}$ and finally:

${|B|}^{{\aleph_{0}}}=\left(\prod_{{w\in W}}{|B_{{|w}}|}^{{\aleph_{0}}}\right)=% \prod_{{w\in T}}{2^{{\aleph_{0}}}}\prod_{{w\in S\setminus T}}{{|B_{{|w}}|}}^{{% \aleph_{0}}}={2^{{\aleph_{0}{|T|}}}}\prod_{{w\in S\setminus T}}{{|B_{{|w}}|}^{% {\aleph_{0}}}}={\left({2{|B_{{|w_{0}}}|}}\right)}^{{\aleph_{0}}}\prod_{{w\in S% \setminus{(T\cup\{w_{0}\})}}}{{|B_{{|w}}|}}=\prod_{{w\in S\setminus T}}{{|B_{{% |w}}|}}\leq|B|\leq{|B|}^{{\aleph_{0}}}$

## 7.33

Let $B$ be a $\kappa$-complete $\kappa$-saturated Boolean algebra. Let $X\subset B$ and $Y=\{b\in B^{+}:\exists x\in X,b\leq x\}\subseteq X$. Let $W\subseteq Y$ be a maximal subset of $Y$ which is an antichain. $B$ is $\kappa$-saturated so $|W|<\kappa$ and $B$ is $\kappa$-complete so $\sum W$ is well-defined.

If there exists $y\in Y$ such that $y\nleq\sum W$ then $0\neq y-\sum W\leq y$. So $y-\sum W\in Y$. For all $w\in W$, $w\cdot\left(y-\sum W\right)\leq{\sum W}\cdot{-\sum W}=0$. This contradicts the maximality of $W$. Hence for all $y\in Y$, $y\leq\sum W$.

If $M$ is a upper bound of $X$ then for all $w\in W\subseteq Y$ there exists $x\in X$ such that $w\leq x\leq M$. So $M$ is an upper bound of $W$ and so $\sum W\leq M$. Moreover, we have seen in the previous paragraph that $\sum W$ is itself an upper bound of $X\subseteq Y$. Hence $\sum X$ is well-defined and $\sum X=\sum W$. Using 7.26, we conclude that $B$ is complete.