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Lattices

We begin the study of lattices and Boolean algebras by generalizing the idea of inequality. Recall that a relation on a set X is a subset of X \times X.

Partially Ordered Sets

We begin the study of lattices and Boolean algebras by generalizing the idea of inequality. Recall that a relation on a set \(X\) is a subset of \(X \times X\). A relation \(P\) on \(X\) is called a partial order of \(X\) if it satisfies the following axioms.

  1. The relation is reflexive: \((a, a) \in P\) for all \(a \in X\).

  2. The relation is antisymmetric: if \((a,b) \in P\) and \((b,a) \in P\), then \(a = b\).

  3. The relation is transitive: if \((a, b) \in P\) and \((b, c) \in P\), then \((a, c) \in P\).

We will usually write \(a \preceq b\) to mean \((a, b) \in P\) unless some symbol is naturally associated with a particular partial order, such as \(a \leq b\) with integers \(a\) and \(b\), or \(A \subset B\) with sets \(A\) and \(B\). A set \(X\) together with a partial order \(\preceq\) is called a partially ordered set, or poset. \(a \preceq b\) \(a\) is less than \(b\)

Example

The set of integers (or rationals or reals) is a poset where \(a \leq b\) has the usual meaning for two integers \(a\) and \(b\) in \({\mathbb Z}\).

Example

Let \(X\) be any set. We will define the power set of \(X\) to be the set of all subsets of \(X\). We denote the power set of \(X\) by \({\mathcal P}(X)\). For example, let \(X = \{ a, b, c \}\). Then \({\mathcal P}(X)\) is the set of all subsets of the set \(\{ a, b, c \}\): \[\begin{aligned}& \emptyset & & \{ a \} & & \{ b \} & & \{ c \} & \\ & \{ a, b \} & & \{ a, c\} & &\{ b, c\} & & \{ a, b, c \}. &\end{aligned}\] On any power set of a set \(X\), set inclusion, \(\subset\), is a partial order. We can represent the order on \(\{ a, b, c \}\) schematically by a diagram such as the one in .

Example

Let \(G\) be a group. The set of subgroups of \(G\) is a poset, where the partial order is set inclusion.

Example

There can be more than one partial order on a particular set. We can form a partial order on \({\mathbb N}\) by \(a \preceq b\) if \(a \mid b\). The relation is certainly reflexive since \(a \mid a\) for all \(a \in {\mathbb N}\). If \(m \mid n\) and \(n \mid m\), then \(m = n\); hence, the relation is also antisymmetric. The relation is transitive, because if \(m \mid n\) and \(n \mid p\), then \(m \mid p\).

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

İndi siz Heç bir hesablama proqramı bu ədədi həll edə bilməz, lakin onun hissələri hesablana bilər. Aşağıdakılardan birini sınayın ya da özünüz yazın.

Özün işlə

Pulsuz hesab hər dərsdə qeydlər əlavə edir, bitirdiyiniz işlərin qeydini saxlayır, həll etdiyiniz problemləri bir yerdə saxlayır və bu səhifə haqqında soruşa biləcəyiniz müəllimi də əlavə edir. Riyaziyyat özü daxil olub-olmadığınıza baxmayaraq hər kəsə açıqdır.

Qeyd Et Giriş

Burada istifadə edilən simvollar

Tam təsviri, şəkli və hər hərfin nə demək olduğunu görmək üçün hər hansı simvolu vurun.

İnsanlar soruşurlar

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Bu səhifənin bəzi hissələri Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Bu, bizim üçün çox böyük bir uğurdur, biz bunu çox istəyirik.

Daha çox Abstract Algebra

GroupsSubgroups, cosets and Lagrange's theoremCyclic groups and permutation groupsHomomorphisms, normal subgroups and quotient groupsRings and fieldsGalois theory: why the quintic has no formula