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

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 $X$.

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 $a_{1},a_{2},...a_{n}\in A$, $\bigcap_{{i<n}}\hat{\{a_{i}\}}={\{Q\in S:\forall i<n,a_{i}\in Q\}}=\hat{\{a_{1},a_{2},...,a_{n}\}}\neq\emptyset$. 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, $F=H$.

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.

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 ${{S\times S}\setminus X}\in V$.

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 $T$.

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:

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.

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.

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 $A_{n}^{-}=\{x\in A_{n}:x<x_{n}\}$ 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

Each $X_{n}^{\pm}$ is finite. If $n_{0}<n_{1}<n_{2}...$ 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 $m_{0}<m_{1}<m_{2}...$ 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 $\omega^{\star}$.

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.

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

Define $X=\{n:f(n)<n\}$. 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 $\{n\in Y:\text{any increasing sequence}\,n<f(n)<f(f(n))<...\,\text{is of finite length}\}$ is not in $D$. Let’s consider the set $Z$ of elements of $Y$ such that there is an infinite increasing sequence $n<f(n)<f(f(n))<...$. 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 $n<f(n)=p<f(f(n))=f(p)<...$. 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 $\{n:f(n)=n\}=\omega\setminus{X\cup Y}\in D$.

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 $D\equiv E$.

(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 $X\in D$.

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.

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 $Y_{n}=X_{n}\setminus\bigcup_{{i<n}}X_{i}$ 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?)

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 $X_{n}$).

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

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

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, $u\cdot(v\oplus w)=u\cdot{(v\cdot-w+-v\cdot w)}=u\cdot v\cdot-w+u\cdot-v\cdot w=u\cdot v\cdot{(-u+-w)}+(-u+-v)\cdot u\cdot w=u\cdot v\cdot{-(u\cdot w)}+-{(u\cdot-v)}\cdot u\cdot w=(u\cdot v)\oplus(u\cdot w)$

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

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: