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Chapter 7: Problem 2

Let \(L=\\{0\\} \cup\left\\{2^{r} 3^{s} \mid r, s=0,1,2,3, \ldots\right\\}\) be ordered by \(\preccurlyeq\) where \(m \preccurlyeq n \Leftrightarrow\left(\exists k \in \mathbb{N}_{0}\right) n=k m\) (i) Prove that \(L\) is a complete, distributive lattice. (ii) Identify the compact (equivalently, finite) elements in \(L\) and in the dual lattice \(L^{2}\). Is \(L\) an algebraic lattice? Is \(L^{\partial}\) an algebraic lattice?

Short Answer

Expert verified
Yes, L is a complete, distributive lattice. All non-zero elements are compact, making both L and its dual algebraic lattices.

Step by step solution

01

Define the Set L in Detail

The set \( L \) is defined as \( \{0\} \cup \{2^r 3^s \mid r, s = 0,1,2,3,\ldots\} \). This includes 0 and all numbers that can be expressed as products of powers of 2 and 3, starting from 2^0 3^0 = 1.
02

Establish Complete Lattice Property

A lattice is complete if every subset has a supremum and infimum. In \( L \), the infimum \( \bigwedge S \) of a subset \( S \) is the greatest common divisor, and the supremum \( \bigvee S \) is the least common multiple. For any pairs of elements \( m = 2^a 3^b \) and \( n = 2^c 3^d \), the lattice operations give infimum \( 2^{\min(a,c)} 3^{\min(b,d)} \) and supremum \( 2^{\max(a,c)} 3^{\max(b,d)} \). Every subset of \( L \) admits these operations, confirming completeness.
03

Prove Distributive Property

A lattice is distributive if for all \( a, b, c \in L \), it holds that \( a \land (b \lor c) = (a \land b) \lor (a \land c) \). Using the gcd and lcm operations given by the powers of 2 and 3, we can verify that the distributive law holds in \( L \). This follows from the identities for gcd and lcm of powers.
04

Identify Compact Elements in L

Compact elements in a lattice are those where, if they are below a supremum of a set, they are below a finite subset of that set. In \( L \), the compact elements are those with only finitely many divisors, which are numbers of the form \( 2^r 3^s \) with finite \( r \) and \( s \). That is, all elements of \( L \) are compact except for \( 0 \).
05

Identify Compact Elements in Dual Lattice L虏

In the dual lattice \( L^\partial \), compact elements are those that are finite in the context of dual ordering. This translates similarly to \( L \) because the set structure remains the same while implication changes. Thus, the compact elements are essentially those numbers that are finite in terms of prime factorization, i.e., \( 2^r 3^s \) with finite \( r \) and \( s \).
06

Determine If L and L虏 Are Algebraic

A lattice is algebraic if every element is a supremum of compact elements (finite elements). Since each element in \( L \) can be expressed by \(2^r 3^s\), which are compact, \( L \) is algebraic. Similarly, \( L^\partial \) is algebraic because the notions of compactness and the lattice structure apply identically.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complete Lattice
A complete lattice is one where every subset has both a supremum (least upper bound) and an infimum (greatest lower bound). In the case of the set \( L \), defined as \( \{0\} \cup \{2^r 3^s \mid r, s = 0,1,2,3,\ldots\} \), this property holds. Every subset of \( L \) can form both of these bounds.
  • The infimum is determined by the greatest common divisor (gcd), ensuring it is the largest number dividing all elements of the subset.
  • The supremum is given by the least common multiple (lcm), which is the smallest number that every element of the subset divides into.

For example, for elements \( m = 2^a 3^b \) and \( n = 2^c 3^d \), the infimum would be \( 2^{\min(a,c)} 3^{\min(b,d)} \) and the supremum \( 2^{\max(a,c)} 3^{\max(b,d)} \). Since these operations are always defined in \( L \), it confirms \( L \) as a complete lattice.
Distributive Lattice
A distributive lattice satisfies a specific property for the meet (\( \land \)) and join (\( \lor \)) operations. Specifically, for any elements \( a, b, c \) in the lattice, it holds that:

\[ a \land (b \lor c) = (a \land b) \lor (a \land c) \]

In the context of \( L \), this property simplifies when elements are expressed in terms of gcd and lcm, as is typical for lattices derived from number factorization.
  • When you calculate gcd and lcm for elements given as \( 2^a 3^b \), the operations naturally respect the distributive identity due to the inherent properties of exponentiation.
  • This distributive nature results from both numeric properties of gcd and lcm aligning with the distributive requirement of lattices.
As such, \( L \) is not only complete but also a distributive lattice, illustrating the elegance of its structure.
Algebraic Lattice
An algebraic lattice is one where every element can be expressed as a supremum of compact elements. Compact elements in a lattice are those that are finitely generated.

In \( L \), which consists of \( 2^r 3^s \) where \( r \) and \( s \) are finite, each element is indeed a supremum of these simple products of powers of 2 and 3.
  • This property ensures that any element in \( L \) can be broken down into more "primitive" components, namely compact elements.
  • Moreover, since \( L \) is a complete lattice, it naturally includes the conditions needed for being algebraic as every element is a combination of its compact counterparts.
Therefore, \( L \) embodies an algebraic structure, supporting both the generality and specificity needed for such classification.
Compact Elements
Compact elements are crucial in the study of lattices, as they help in understanding its generative structure. In \( L \), compact elements arise from the finite exponentiation of 2 and 3, i.e., elements expressed as \( 2^r 3^s \).

  • These elements showcase their compact nature as they have only a finite number of divisors compared to potential infinite divisors one might imagine in other cases.
  • An element is termed compact if whenever it lies below a join (supremum) of some subset, it must lie below the join (supremum) of some finite subset of this subset.
In the dual lattice, \( L^\partial \), the concept of compactness translates similarly, emphasizing finitely generable elements based on duality principles. Compact elements present foundational building blocks for \( L \) as well as \( L^\partial \), ensuring both retain their algebraic characteristics.

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Most popular questions from this chapter

Let \(L\) be, an algebraic lattice and \(K\) a non-empty subset of \(L\) such that \(\bigvee_{L} S\) and \(\wedge_{L} S\) belong to \(K\) for every non-empty subset \(S\) of \(K .\) Show that \(K\) is an algebraic lattice. [Hint. Represent \(L\) as a topped algebraic \(\bigcap\) -structure on some set \(X\) and show that \(K\) is also a topped algebraic \(\bigcap\) -structure on some subset \(Y\) of \(X .\) Deduce that the interval \([x, y]:=\\{z \in L \mid x \leqslant z \leqslant y\\}\) is an alge braic lattice for all \(x, y \in L\) with \(x \leqslant y\).

Let \(P\) be an ordered set and let \(S \subseteq P .\) Prove that if the joins on the right-hand side exist then \(V S\) exists and is given by $$ V S=\bigcup\\{\bigvee F \mid \varnothing \neq F \subseteq S, F \text { finite }\\}. $$

Let \(L\) be a complete lattice. Prove that the following statements are equivalent: (i) \(K(L)=L\); (ii) every directed subset of \(L\) has a greatest element; (iii) \(L\) satisfies \((\mathrm{ACC})\).

Let \(P, Q\) and \(R\) be ordered sets and let \(\varphi: P \rightarrow Q\) and \(\psi: Q \rightarrow R\) be order-preserving maps. Prove that if \(\varphi\) and \(\psi\) have upper adjoints \(\varphi^{\\#}\) and \(\psi^{\\#}\), respectively, then the composite \(\psi \circ \varphi\) has upper adjoint \(\varphi^{\dagger} \circ \psi^{\|}\) (so that \(\left(\psi \circ \varphi, \varphi^{\dagger} o \psi^{\\#}\right)\) is a Galois connection).

Recall that a quasi-order on a set \(P\) is a binary relation on \(P\) which is reflexive and transitive. Let \(P\) be a set and assume that \(\leqslant_{1}\) and \(\leqslant_{2}\) are quasi-orders on \(P\) such that \(x \leqslant 1 y\) and \(x \leqslant 2 y\) imply \(x=y .\) Define, for a subset \(Y\) of \(P\), $$ \begin{aligned} &Y^{\triangleright}:=\\{x \in P \mid(\forall y)(y \leqslant 1 x \Rightarrow y \notin Y)\\} ,\\\ &Y^{\triangleleft}:=\\{x \in P \mid(\forall y)(y \leqslant 2 x \Rightarrow y \notin Y)\\}. \end{aligned} $$ A subset \(Y\) of \(P\) is called stable if \(Y^{\triangleright \triangleleft}=Y\); denote the set of all stable subsets of \(P\) by \(\mathcal{S}(P)\). (i) Verify that the pair \((D, \triangleleft)\) sets up a Galois connection be tween the \(\leqslant 1\) -down-sets of \(P\) ordered by \(\subseteq\) and the \(\leqslant_{2}\) -downsets of \(P\) ordered by \(\supseteq\). (Down-sets of a quasi-ordered set are defined in just the same way as down-sets of an ordered set.) (ii) Verify that \(\langle\mathcal{S}(P) ; \subseteq\rangle\) is a lattice in which meet and join are given by \(Y \wedge Z=Y \cap Z\) and \(Y \vee Z=\left(Y^{\triangleright} \cap Z^{\triangleright}\right)^{\triangleleft}\). (iii) Let \(P\) be a 3-element set \(\\{a, b, c\\}\) with \leqslanti defined by \(x \leqslant_{1} y\) if and only if \(x=y\) or \(x=a\) and \(y=b\) and \(\leqslant 2\) defined by \(x \leqslant_{2} y\) if and only if \(x=y\) or \(x=a\) and \(y=c\). Identify the stable subsets of \(P\) and prove that \(\langle S(P) ; \subseteq\rangle \cong \mathbf{N}_{5}\) (iv) Let \(P\) be a 6-element set with elemeats \(a_{1}, a_{2}, \ldots, a_{6}\) and define \(\leqslant_{1}\) and \(\leqslant_{2}\) by $$ \begin{aligned} &x \leqslant 1 y \Longleftrightarrow x=y \text { or }\\{x, y\\} \in\left\\{\left\\{a_{1}, a_{2}\right\\},\left\\{a_{3}, a_{4}\right\\},\left\\{a_{5}, a_{6}\right\\}\right\\} \\ &x \leqslant_{2} y \Longleftrightarrow x=y \text { or }\\{x, y\\} \in\left\\{\left\\{a_{2}, a_{3}\right\\},\left\\{a_{4}, a_{5}\right\\},\left\\{a_{6}, a_{1}\right\\}\right\\} \end{aligned} $$ Find the stable subsets 贸f \(P\) and prove that \(\langle\mathcal{S}(P) ; \subseteq\rangle \cong \mathrm{M}_{3}\). (This exercise hints at the concrete representation for arbitrary bounded lattices, due to A. Urquhart, which in the finite case generalizes Birkhoff's representation. In the distributive case, \(\leqslant_{1}\) and \(\leqslant 2\) are respectively \(\leqslant\) and \(\geqslant\), for an order \(\left.\leqslant .\right)\)
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