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\newtheorem*{first}{The First Homomorphism Theorem}
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\begin{document}

\title{Ordered Algebraic Structures \\ 
Synopsis of a course at OU, winter semesters 2017/2018 and 2018/2019}
\author{Pasha Zusmanovich}
\address{}
\email{pasha.zusmanovich@gmail.com}
\date{last modified December 28, 2018}

\maketitle


%{\small 
%Classes at 2017/2018 lasted 1.5 hours, classes at 2018/2019 last 3 hours 
%(``lecture'' + ``tutorials''). Thus the synopsis is gradually converted to the 
%3-hours format}.


\section{
Algebraic structures, substructures, automorphisms. Orders \texorpdfstring{\\}{}
(October 4, 2018)
}\label{sec-2} % class 1

{\small
\noindent
Literature: 
\cite[Chap.I,\S 1, Chap.II,\S\S 1,2]{main}; 
\cite[\S\S 1.1,1.2,2.1]{c.bergman}; 
\cite[\S\S 1.4,1.5,2.2,2.3]{malcev}.
}

\medskip

The basic structures in mathematics are: algebraic, topological, and ordered
structures. Combinations of these leads to various branches of mathematics.

Definition of an algebraic structure.
Notion of a substructure of an algebraic structure.

Definition of order; partial and total (= linear) orders. 

Examples: $(\mathbb N, \le)$, $(\mathbb Z, \le)$, 
$P(X)$ (set of subsets of the given set $X$). The latter order is total if
an only if $|X| \le 1$.

Cartesian product of orders: if $(X_i, \le)$ are orders, $i=1,\dots,n$, then the
order $X_1 \times \dots \times X_n$ is defined as 
$(x_1, \dots, x_n) \le (y_1, \dots, y_n)$ iff $x_i \le y_i$ for any 
$i=1,\dots,n$. This order is not total except degenerate cases ($|X_i| \le 1$).
For example, complex numbers, considered as pairs of real numbers, may be 
ordered this way.

\medskip

Examples of algebraic structures:
\begin{enumerate}[\upshape(i)]
\item
$(\mathbb N, \mapsto)$, where $\mapsto$ is the unary function $n \mapsto n+1$.
All substructures are of the form \newline $\{n,n+1,n+2,\dots\}$ for some n.
\item
A dynamical system: $(X, f)$, for unary $f: X \to X$.
\item
$(C^\infty(\mathbb R), \frac{d}{d\,x}, +, \cdot)$, where $+$ and $\cdot$ are
pointwise addition and multiplication of functions.
Examples of substructures: polynomials; functions of the form 
$a_1 e^{b_1 x} + \dots + a_n e^{b_n x}$, where $a_i,b_i \in \mathbb R$.
\item
$(M_n(\mathbb R), t)$, where $t(A,B,C) = ABC$.
\item
A ``circular'' variant of (i): $(\{1,2,3,4\}, \mapsto)$, where 
$1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 1$. There are no proper substructures.
\item
Small structures with operations of small arity can be given by 
``multiplication'' tables.
\end{enumerate}

\begin{theorem}
The intersection of any number of substructures of an algebraic structure is a 
substructure.
\end{theorem}

Notions of homomorphism, isomorphism, automorphism.

Example of isomorphism: $(\mathbb R, +) \simeq (\mathbb R_{>0}, \cdot)$,
where isomorphism is provided by $x \mapsto e^x$.

\begin{theorem}
Automorphisms of an algebraic system form a group.
\end{theorem}

Example: automorphisms of the structure from Example (v) form the cyclic group
$\mathbb Z/4\mathbb Z$.




\section{Congruences. Homomorphism theorems (October 11, 2018)} % class 2

{\small
\noindent
Literature: 
\cite[Chap.II, \S\S 5,6]{main}; 
\cite[\S\S 1.5,3.1]{c.bergman}; 
\cite[\S\S 2.4,3.3,4.1]{malcev}.
}

\medskip

Refresher: equivalence relation, equivalence classes.

Definition of congruence. Congruences for groups amount to normal subgroups,
and congruences for rings amount to ideals.

Any algebraic structure $X$ has trivial congruences: the minimal one -- the 
diagonal $\Delta(X) = \set{(x,x)}{x \in X}$, and the maximal one -- the whole 
Cartesian product $X \times X$.

\begin{definition}
If $\varphi: X \to Y$ is a homomorphism of algebraic structures (of the same
signature), then its kernel, denoted by $\Ker \varphi$, is defined as 
$\set{(a,b) \in X}{\varphi(a) = \varphi(b)}$.
\end{definition}

\begin{theorem}
A kernel of homomorphism is a congruence.
\end{theorem}

\begin{example}
The only nontrivial congruence of the structure (v) from Class \ref{sec-2} is:
\newline $\Delta(x) \cup \{(1,3),(3,1),(2,4),(4,2)\}$.
\end{example}





\medskip

\begin{first}
If $\varphi: X \to Y$ is a surjective homomorphism of algebraic structures, then
$X/\Ker\varphi \simeq Y$.
\end{first}

\begin{definition}
If $\alpha \subseteq \beta$ are congruences on an algebraic structure $X$, then
$$
\beta / \alpha \overset{df}= 
\set{(x/\alpha, y/\alpha) \in X/\alpha \times X/\alpha}{(x,y) \in \beta} .
$$
\end{definition}

\begin{lemma}
$\beta / \alpha$ is a congruence on $X / \alpha$.
\end{lemma}

\begin{second}
If $\alpha \subseteq \beta$ are congruences on an algebraic structure $X$,
then $(X/\alpha) / (\beta / \alpha) \simeq X/\beta$.
\end{second}

\begin{proof}
Establish a map $X/\alpha \to X/\beta$, and use the First Homomorphism Theorem.
\end{proof}

\begin{lemma}
If $X$ is a substructure of, and $\alpha$ is a congruence on an algebraic 
structure $Y$, then $\alpha \cap (X \times X)$ is a congruence on $X$.
\end{lemma}

\begin{third}
If $X$ is a substructure of, and $\alpha$ is a congruence on an algebraic 
structure $Y$, then $X / (\alpha \cap (X \times X))$ is isomorphic to a 
substructure of $Y/\alpha$.
\end{third}

\begin{proof}
Establish a map $X / (\alpha \cap (X \times X)) \to Y/\alpha$, prove that it is
injective, and use the First Homomorphism Theorem.
\end{proof}






\section{Lattices (October 18, 2018)} % class 3

{\small
\noindent
Literature: 
\cite[Chap.I,\S 1, Chap.II,\S 5]{main}; 
\cite[\S\S 1.4,2.1]{c.bergman}; 
\cite[\S\S 2.3,5.1]{malcev};
Wikipedia: \emph{Lattice (order)}.
}

\medskip

Notions of supremum and infimum of a subset of an ordered set.

\begin{definition}
A lattice is an ordered set in which any two elements have supremum and infimum
(called join and meet, respectively).
\end{definition}

\begin{definition}
A lattice is an algebraic structure of the form $(X, \wedge, \vee)$, where 
$\wedge$ and $\vee$ are binary operations satisfying the following axioms:
\begin{enumerate}
\item both $\wedge$ and $\vee$ are commutative and associative;
\item (absorption) $a \vee (a \wedge b) = a$, $a \wedge (a \vee b) = a$.
\end{enumerate}
\end{definition}

Equivalence of these two definitions: 
1 $\Rightarrow$ 2: $a \vee b = sup(a,b)$, $a \wedge b = inf(a,b)$.
\\
2 $\Rightarrow$ 1: $a \le b$ iff $a = a \vee b$ iff $b = a \wedge b$.

Idempotency in lattices: $a \wedge a = a$, $a \vee a = a$. Follows from 
absorption, for example: $a \vee a = a \vee (a \wedge (a \vee a)) = a$.

\bigskip

Intersection of congruences on an algebraic structure is a congruence
(had to be earlier, when talking about congruences).

Notions of substructure of and congruence on algebraic structure generated by a
subset (had to be earlier, when talking about substructures and congruences
in arbitrary algebraic structures).

\bigskip

Examples: in an arbitrary lattice, every element generates an one-element
sublattice. Every two element generate either two-element totally ordered
sublattice, or $4$-element ``diamond'' sublattice $D_4$, depending whether they
are comparable or not.

Any lattice consisting of $\le 4$ elements isomorphic to one of the 5 lattices:
a linear order $L_1$, $L_2$, $L_3$, $L_4$ (consisting of $1$, $2$, $3$, $4$ 
elements respectively), or $D_4$.

Hasse diagram of a lattice.

Examples: $P(X)$ (the set of all subsets of a set $X$) forms a lattice; 
for any $A \subseteq X$, $P(A)$ is a sublattice.
The lattice $(\mathbb N, |)$ ($|$ means ``divides'') is isomorphic (through the
prime numbers decomposition) to the countable direct power of the lattice $(\mathbb N, \le)$.

\bigskip

Substructures of and congruences on a given algebraic structure form lattices.

Example: the lattice of substructures of the ``circular'' structure from 
Class \ref{sec-2}, Example (v) is isomorphic to the one-element lattice $L_1$,
and the lattice of its congruences is isomorphic to $L_3$.

Lattice of congruences of a totally ordered set.





\section{Distributive and modular lattices (October 25, 2018)} % class 4

{\small
\noindent
Literature:
\cite[Chap.I,\S 3]{main}; \cite[\S 2.2]{c.bergman}; \cite[\S 5.2]{malcev}.
}

\medskip

%Examples:
%1. It does matter which operations we consider when talking about substructures.
%For example, the sublattice of the lattice of all subsets of the given set
%$X$ generated by two subsets will consists of $2$ and $4$ elements, but if we
%add another unary operation -- complement, these substructures will be, 
%generally, much larger.

Exercise: Find lattices of sublattices of and congruences on the $4$-element
diamond lattice $D_4$.

Answer: Sublattices form a certain $12$-element lattice, $Con(D_4) \simeq D_4$.

\medskip

The question about congruences on the lattice $P(X)$ (for arbitrary $X$) is a
difficult one.

Dual lattice.

\begin{definition}\label{def-1}
A lattice $L$ is called distributive if one of the following three equivalent
condition holds:
\begin{enumerate}[\upshape(i)]
\item Distributivity of $\vee$ with respect to $\wedge$: 
$x \wedge (y \vee z) = (x \wedge y) \vee (x \wedge z)$ for any $x,y,z \in L$;
\item Distributivity of $\wedge$ with respect to $\vee$: 
$x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z)$ for any $x,y,z \in L$;
\item 
$x \wedge (y \vee z) \le (x \wedge y) \vee (x \wedge z)$ for any $x,y,z \in L$;
\item 
$x \vee (y \wedge z) \ge (x \vee y) \wedge (x \vee z)$ for any $x,y,z \in L$.
\end{enumerate}
\end{definition}

\begin{lemma}\label{fact-1}
In any lattice $L$, the following holds for any $x,y,z \in L$:
\begin{enumerate}[\upshape(i)]
\item $(x \wedge y) \vee (x \wedge z) \le x \wedge (y \vee z)$
\item $(x \vee y) \wedge (x \vee z) \ge x \vee (y \wedge z)$
\end{enumerate}
\end{lemma}

\begin{proof}
(i) Since $x \wedge y \le x$, we have 
$(x \wedge y) \vee (x \wedge z) \le x \vee (x \wedge z) = x$ (by absorption).
Since $x \wedge y \le y$ and $x \wedge z \le z$, we have 
$(x \wedge y) \vee (x \wedge z) \le y \vee z$. Hence
$(x \wedge y) \vee (x \wedge z) \le x \wedge (y \vee z)$, as required.

(ii) By duality.
\end{proof}

\begin{proof}[Proof of equivalences in Definition \ref{def-1}]
(i) $\Rightarrow$ (ii)
\begin{align*}
x \vee (y \wedge z) 
&= (x \vee (x \wedge z)) \vee (y \wedge z) \text{ (by absorption)} \\
&= x \vee ((x \wedge z) \vee (y \wedge z)) \text{ (by associativity)} \\
&= x \vee ((z \wedge x) \vee (z \wedge y)) \text{ (by commutativity)} \\
&= x \vee (z \wedge (x \vee y)) \text{ (by (i))}     \\
&= x \vee ((x \vee y) \wedge z) \text{ (by commutativity)}     \\
&= (x \wedge (x \vee y)) \vee ((x \vee y) \wedge z) \text{ (by absorption)} \\
&= ((x \vee y) \wedge x) \vee ((x \vee y) \wedge z) \text{ (by commutativity)} \\
&= (x \vee y) \wedge (x \vee z) \text{ (by (i))}.
\end{align*}

(ii) $\Rightarrow$ (i) By duality.

(i) $\Leftrightarrow$ (iii) follows from Fact \ref{fact-1}(i).

(ii) $\Leftrightarrow$ (iv) follows from Fact \ref{fact-1}(ii) (or by duality).
\end{proof}

Example of distributive lattices: linear orders, $P(X)$. 

Example of non-distributive lattice: $M_5$.

\begin{definition}
A lattice is called modular if one of the following equivalent conditions holds:
\begin{enumerate}[\upshape(i)]
\item $x \le y \Rightarrow y \wedge (x \vee z) = x \vee (y \wedge z)$;
\item $(x \wedge y) \vee (z \wedge y) = ((x \wedge y) \vee z) \wedge y$.
\end{enumerate}
\end{definition}

\begin{proof}[Proof of equivalence in this definition]
(i) $\Rightarrow$ (ii)
We have $x \wedge y \le y$, hence 
$y \wedge ((x \wedge y) \vee z) = (x \wedge y)\vee (y \wedge z)$, what, 
up to commutativity, is (ii).

(ii) $\Rightarrow$ (i) If $x \le y$, then $x = x \wedge y$, and the identity
(ii) becomes $x \vee (z \wedge y) = ((x \vee z) \wedge y$, what, up to
commutativity, is implication in (i).
\end{proof}

\begin{theorem}
Any distributive lattice is modular.
\end{theorem}

\begin{proof}
If $x \le y$, then $x \vee y = y$, and 
$y \wedge (x \vee z) = (x \vee y) \wedge (x \vee z) = x \vee (y \wedge z)$
(by distributivity).
\end{proof}

When checking a lattice for distributivity or modularity, it is enough to 
consider triples of elements which are all different, and not contain 
$0$ and $1$ (the minimal and maximal elements), if they exist.

Exercise: check that the lattice $M_5$ is modular, and $N_5$ is not distributive
and not modular.

\begin{theorem}\label{th-vect}
Let $V$ be a vector space. Then  the lattice of subspaces of $V$ is modular.
\end{theorem}

Question: whether it is distributive?





\section{Distributive and modular lattices (cont.). Complemented lattices. 
Boolean algebras (November 1, 2018)} % class 5

{\small
\noindent
Literature: 
\cite[Chap.I,\S 3, Chap.IV,\S 1]{main}; \cite[\S 2.2]{c.bergman}; \cite[\S 5.2]{malcev}.
}

\medskip

\begin{theorem}
Let $V$ be a vector space. Then the lattice of subspaces of $V$ is distributive
iff $\dim V = 0$ or $1$.
\end{theorem}

\begin{proof}
The cases of $\dim V = 0$ or $1$ are obvious. Assume $\dim V \ge 2$.

Case 1. The characteristic of the ground field is $\ne 2$. Choose
two linearly independent vectors $u$ and $v$. Then the three one-dimensional 
vector spaces $\langle u \rangle$, $\langle u + v \rangle$, 
$\langle u - v \rangle$ provide counterexample to the distributivity:
$$
(\langle u-v \rangle + \langle u+v \rangle) \cap \langle u \rangle =
\langle u,v \rangle \cap \langle u \rangle = \langle u \rangle ,
$$
but
$$
\langle u-v \rangle \cap \langle u \rangle + 
\langle u+v \rangle \cap \langle u \rangle = 0 + 0 = 0 .
$$

Case 2. The characteristic of the ground field is $2$. Over $GF(2)$, the
lattices of subspaces of a $2$-dimensional space is isomorphic to $M_5$ which
is not distributive. Since enlargement of the vector space, and enlargement of
the ground field lead to a bigger lattice, it will be also not distributive,
and we are done.
\end{proof}

Dedekind's and Birkhoff's theorems about characterization of modular and
distributive lattices in terms of (not) containment of $N_5$ and $M_5$.

Proof of the Dedekind theorem.

Another proof of Theorem \ref{th-vect} using the Dedekind theorem.

\medskip

Complemented lattices: definition.

Exercise: Which of the following lattices are complemented: $P(X)$, total order,
$M_5$, $N_5$. (the latter two lattices show that complement does not have to be
unique).

\begin{definition} 
A Boolean algebra is a distributive complemented lattice.
\end{definition}

\begin{definition} 
A Boolean algebra is an algebraic system with two binary operations $\vee$ and
$\wedge$, one unary operation $\neg$, and two distinguished elements $0$ and 
$1$, satisfying the (highly redundant) system of axioms:
\begin{enumerate}
\item $\neg 0 = 1$, $\neg 1 = 0$;
\item $\neg\neg x = x$;
\item $0 \vee x = x$, $0 \wedge x = 0$, $1 \vee x = 1$, $1 \wedge x = x$;
\item $x \wedge x = x$, $x \vee x = x$;
\item $\wedge$ and $\vee$ are commutative, associative, and distributive with 
      respect to each other;
\item (de Morgan laws) 
      $\neg (x \wedge y) = (\neg x) \vee (\neg y)$,
      $\neg (x \vee y) = (\neg x) \wedge (\neg y)$. 
\end{enumerate}
\end{definition}

Equivalence of two definitions of Boolean algebras (in a complemented 
distributive lattice, complements are unique).

Significance of Boolean algebras.





\section{Boolean algebras (cont). Direct product (November 8, 2018)} % class 6

{\small
\noindent
Literature: 
\cite[Chap.II,\S 7, Chap.IV,\S 1]{main}; 
\cite[\S\S 1.3,3.2]{c.bergman}; 
\cite[\S 2.5]{malcev}.
}

\medskip

Examples of Boolean algebras: two-element Boolean algebra $\mathbf{2}$, P(X).

Exercise: find a 3-element Boolean algebra.

Answer: such Boolean algebras do not exist, because each 3-element lattice
is a total order, and

\begin{proposition}
A Boolean algebra is a total order iff it is isomorphic to $\mathbf{2}$.
\end{proposition}

Direct product of algebraic systems. In general, unlike in the group case, 
factors are not necessary subsystems in their direct product.
Properties of direct product: commutativity and associativity.

Direct product of linear orders is not a linear order. Direct product of Boolean
algebras is a Boolean algebra.

\medskip

Homomorphism of direct product to factors.

Notion of directly indecomposable algebraic system.

Examples: 4-element ``diamond'' decomposes as $\mathbf{2} \times \mathbf{2}$;
linear orders are directly indecomposable; 
simple algebraic systems are directly indecomposable.

$P(X) \simeq \mathbf{2}^X$.

Notion of restriction $B|a$ for a Boolean algebra $B$ and $a\in B$.

Homomorphism $B \to B|a$.

\begin{lemma}
For any Boolean algebra $B$, and any $a \in B$, $B \simeq B|a \times B|\neg a$.
\end{lemma}






\section{The Stone theorems. Subdirect irreducibility (November 15, 2018)} % class 7

{\small
\noindent
Literature: 
\cite[Chap.II,\S\S 7,8, Chap.IV,\S 1]{main}; 
\cite[\S 3.3]{c.bergman};
\cite[\S 5.2]{malcev}.
}

\medskip

\begin{corollary}\label{cor-b}
A Boolean algebra is directly indecomposable iff it is isomorphic to 
$\mathbf{2}$.
\end{corollary}

\begin{theorem}[``The Little Stone Theorem'']
Any finite Boolean algebra is isomorphic to $P(X)$ for a finite set $X$.
\end{theorem}

Proof is by induction, using Corollary \ref{cor-b} and the fact that 
$P(X) \simeq \mathbf{2}^X$.

\begin{corollary}
For two finite Boolean algebras $B_1$ and $B_2$, 
$B_1 \simeq B_2$ iff $|B_1| = |B_2|$.
\end{corollary}

\begin{theorem}[``The Big Stone Theorem'']\label{th-st}
Any Boolean algebra is a subalgebra of $P(X)$ for some set $X$.
\end{theorem}

An example of an (infinite) Boolean algebra not isomorphic to $P(X)$: the set
of all finite and all cofinite subsets of an infinite set $X$ (to finish the
proof is a Homework).

Notion of subdirect product.

An equivalent formulation of the Big Stone Theorem: any Boolean algebra is a 
subdirect power of $\mathbf{2}$.

Notion of subdirect irreducibility of an algebraic structure.

Examples of subdirectly irreducible algebraic structures: $2$-element structures,
simple structures.

A vector space is subdirectly irreducible iff it is of dimension $0$ or $1$.

A finite abelian group is subdirectly irreducible iff it is isomorphic to a
cyclic group of a prime power order.







\section{Proof of the Big Stone Theorem. Boolean rings} % class 8

{\small
\noindent
Literature: 
\cite[Chap.II,\S\S 6,8, Chap.IV,\S\S 1,2]{main}; \cite[\S\S 2.1,3.1,3.3,3.4]{c.bergman}.
}

\medskip

Notion of the interval $[a,b]$ in a lattice.

Theorem: for any algebraic structure $A$, and any $\theta \in Con(A)$,
$[\theta,\nabla_A] \simeq Con(A/\theta)$. Corollary: a quotient of an algebraic
structure by a maximal proper congruence is simple.

Criterion of subdirect irreducibility: an algebraic structure $A$ is subdirectly
irreducible iff there is a smallest element in $Con(A) \backslash \{\Delta_A\}$.
Corollary: any simple algebraic structure is subdirectly irreducible.

\begin{theorem}[Birkhoff]
Any algebraic structure is a substructure of a direct product of subdirectly 
irreducible structures.
\end{theorem}

Zorn's lemma.

Finish of the proof of the Stone theorem.

\medskip

Boolean rings. Correspondence Boolean rings $\leftrightarrow$ Boolean algebras.

Exercise: Which Boolean rings are fields? Answer: $GL(2)$.

Finite Boolean rings are direct sums of copies of $GF(2)$ (follows from Stone's
theorem).






\section{Ideals, filters and ultrafilters in Boolean algebras} % class 9

{\small
\noindent
Literature: 
\cite[Chap.IV,\S 3]{main}, \cite[\S 8.1]{malcev}.
}

Ideals in Boolean rings lead to ideals in Boolean algebras.

Definition of ideal and filter in a Boolean algebra, their duality.

Examples of filters: cofinite filter in $P(X)$, principal ultrafilter.

Ultrafilters as maximal proper filters.

A filter $F$ in a Boolean algebra $B$ is an ultrafilter iff for any $a\in B$,
either $a\in F$, or $\neg a \in F$.

Description of filters on finite Boolean algebras.



\section{Stone's duality} % class 10

{\small
\noindent
Literature: 
\cite[Chap.IV,\S 4]{main}.
}

Discussion of homeworks.

Homework 6: to prove that the lattice of normal subgroups of a group is modular is more-or-less routine task, but to describe groups for which
this lattice is distributive, is more like a research problem (for example,
for a group which is a direct product of $n$ simple groups this lattice is
isomorphic to lattice of subsets of an $n$-element set and hence is 
distributive).


%Discussion of homework 10: clopen sets in any topology form a Boolean algebra.

\medskip

Boolean (= Stone) topological spaces. Correspondence between Boolean algebras 
and Boolean spaces.



\begin{lemma}\label{lemma-i}
Let $B$ be a Boolean algebra, $X$ a subset of $B$. Then the ideal of $B$
generated by $X$ (= the minimal ideal of $B$ containing $X$) coincides with
$$
\set{b\in B}{b \le x_1 \vee \dots \vee x_n, x_1, \dots, x_n \in X} \cup \{0\} .
$$
\end{lemma}

For an (easy) proof, see \cite[Chap. IV, Lemma 3.9(a)]{main}.

\begin{proof}[Proof that for a Boolean algebra $B$, $B^*$ is compact]
Let $\set{N_a}{a \in X}$ be a cover of $B^*$. Consider the set $\mathcal I$ of 
all proper ideals of $B$ containing $X$. 

Case 1. $\mathcal I = \varnothing$. Then the ideal generated by $X$ coincides
with $B$, and by Lemma \ref{lemma-i}, $1 = x_1 \vee \dots \vee x_n$ for some 
$x_1, \dots, x_n \in X$. Let $U \in B^*$ (i.e., $U$ is an ultrafilter of $B$).
Since $1 \in U$, we have $x_i \in U$ for some $1 \le i \le n$, i.e. 
$U \in N_{x_i}$. Hence $N_{x_1}, \dots, N_{x_n}$ is a finite (sub)cover of 
$B^*$.

Case 2. $\mathcal I \ne \varnothing$. Then by Zorn's lemma, $X$ contained in 
some maximal ideal $I$ of $B$. Then $U = \neg I$ is an ultrafilter, and
$U \cap I = \varnothing$. But then for any $a \in X$, we have $a \in I$, hence
$a \notin U$, and $U \notin N_a$, a contradiction with the fact that
$\set{N_a}{a \in X}$ is a cover of $B^*$
\end{proof}

\begin{proof}[Proof that the map $B \to B^{**}$, $b \mapsto N_b$, is injective]
Let $a,b\in B$, $a \ne b$. Then $(a \vee b) \wedge \neg(a \wedge b) \ne 0$, and
there is an ultrafilter $U$ on $B$ such that 
$(a \vee b) \wedge \neg(a \wedge b) \in U$. But since 
$a \vee b \ge (a \vee b) \wedge \neg(a \wedge b)$, and $a \vee b \in U$,
and hence $a \in U$ or $b \in U$. Similarly, 
$\neg(a \wedge b) = \neg a \vee \neg b \in U$, and $\neg a \in U$ or 
$\neg b \in U$, what is equivalent to $a \notin U$ or $b \notin U$. Thus, 
exactly one of $a$, $b$ belongs to $U$, i.e. $U$ lies in exactly one of $N_a$,
$N_b$, so $N_a \ne N_b$.
\end{proof}

\begin{proof}[Proof that the map $B \to B^{**}$, $b \mapsto N_b$, is surjective]
Let $N$ be a clopen subset of $B^*$. Then $N$ is a union of a number of $N_a$'s.
But since $N$ is a closed subset of a compact space, $N$ is compact, and hence
is a union of a finite number of $N_a$'s, say, 
$N = N_{a_1} \cup \dots \cup N_{a_n} = N_{a_1 \vee \dots \vee a_n}$ (by the 
lemma proved at the previous class).
\end{proof}





\begin{thebibliography}{BS}

\bibitem[B]{c.bergman} C. Bergman, \emph{Universal Algebra}, CRC Press, 2012.

\bibitem[BS]{main} 
S. Burris and H.P. Sankappanavar, \emph{A Course in Universal Algebra}, 
The Millennium Edition.

\bibitem[M]{malcev} A.I. Mal'cev, \emph{Algebraic Systems}, Springer, 1973.

\end{thebibliography}

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