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Chapter 8: Problem 16

Construct the least solutions to the following equations in the CPO \(P=\wp(X)\) (i) \(S=S \cup T \quad(T\) fixed \()\) (ii) (with \(X=\mathbb{N}) S=S \cup\\{1\\} \cup\\{n+2 \mid n \in S\\}\) (iii) (with \(X=\mathbb{N}^{2}\) ) $$ S=\\{(n, n) \mid n \in \mathbb{N}\\} \cup\\{(n, m+1) \mid(n, m) \in S\\} $$ (iv) (with \(X\) a finite group) \(S=T \cup S \cup S \cdot S \quad(T\) fixed). [Hint. Find a suitable order-preserving map \(F\) of which the required set is to be the least fixpoint, guess a formula for \(F^{n}(\perp)\) and prove by induction that it works, and finally verify that \(\bigcup F^{n}(\perp)\) is a fixpoint.] (See Theorem \(8.15(\mathrm{i}) .)\)

Short Answer

Expert verified
The least solutions are: (i) \(T\), (ii) \{1, 3, 5, 7, \ldots\}\, (iii) \{(n, m) \mid m \geq n\}\, (iv) the subgroup generated by \(T\).

Step by step solution

01

Understanding the Problem

We need to construct the least solution for each given equation in the complete partial order (CPO) of subsets of a set \(X\). This involves finding the smallest fixpoint (if it exists) of a function. The hint suggests employing an order-preserving map \(F\) and iterating from the least element \(\perp\) of the CPO to find \(\bigcup F^n(\perp)\).
02

Item (i) Analysis

For \(S = S \cup T\), consider the map \(F(S) = S \cup T\). The least element \(\perp\) in \(\wp(X)\) is the empty set. We need to find the sequential applications of \(F\) starting from \(\perp\).
03

Item (i) Iteration Steps

Applying \(F\), we get \(F^0(\perp) = \perp = \emptyset\), \(F^1(\perp) = \emptyset \cup T = T\), and \(F^2(T) = T \cup T = T\). Hence, the sequence converges at \(F^1\) and the least solution is \(T\).
04

Construct a Formula for (i)

The least fixpoint \(\bigcup F^n(\perp) = T\) as it stabilizes at \(T\) without further changes. Since \(F^2 = F^1\), it confirms \(T\) as the least fixpoint.
05

Item (ii) Analysis

For \(S = S \cup \{1\} \cup \{n+2 \mid n \in S\}\), we define \(F(S)=S \cup \{1\} \cup \{n+2 \mid n \in S\}\). The base element is \(\emptyset\), and we begin iterating F starting with \(F^0(\perp) = \emptyset\).
06

Item (ii) Iteration Steps

\(F^1(\emptyset) = \{1\}\), \(F^2(\{1\}) = \{1, 3\}\), \(F^3(\{1, 3\}) = \{1, 3, 5\}\), etc., generating \(\{1, 3, 5, 7, \ldots\}\). Thus, \(F^n(\emptyset) = \{1, 3, 5, \ldots, 2n-1\}\).
07

Construct a Formula for (ii)

The limit \(\bigcup F^n(\perp) = \{1, 3, 5, 7, \ldots\}\), confirming it as the least fixpoint for (ii). The formula derived by induction works as each step adheres to the sequence of generating next odd integers.
08

Item (iii) Analysis

For \(S=\{(n, n) \mid n \in \mathbb{N}\} \cup \{(n, m+1) \mid (n, m) \in S\} \), we let \(F\) map subsets of \(\mathbb{N}^2\) starting from \(\emptyset\) iterating through \(F\).
09

Item (iii) Iteration Steps

Starting at \(F^0(\perp) = \emptyset\), after applying \(F\): \(F^1(\emptyset) = \{(n, n) \mid n \in \mathbb{N}\}\); then, each iteration adds one more element to the second component of the pairs, forming all pairs \((n, m)\) with \(m \geq n\).
10

Construct a Formula for (iii)

\(\bigcup F^n(\perp) = \{(n, m) \mid m \geq n\}\), capturing the desired pairs by transitive closure of pair closures up the diagonal.
11

Item (iv) Analysis

For \(S = T \cup S \cup S \cdot S\) with \(T\) fixed and \(\cdot\) the group operation, let \(F(S) = T \cup S \cup S \cdot S\). We apply \(F\) iteratively.
12

Item (iv) Iteration Steps

Starting at \(F^0(\perp) = \emptyset, F^1(\emptyset) = T, F^2(T) = T \cup T \cdot T\), iterating continues to extend including closed powers and products by group axioms forming the subgroup generated by elements of \(T\).
13

Construct a Formula for (iv)

The least fixpoint \(\bigcup F^n(\perp)\) equals the subgroup generated by \(T\), derived as it captures closure properties of groups, confirming initial closure bases and products by definition.

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Key Concepts

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

Least Fixpoint
In mathematics, particularly in the context of complete partial orders (CPO), the term "least fixpoint" has a significant role. A fixpoint of a function is a point that is mapped to itself by the function. The least fixpoint, therefore, is the smallest point (or subset) that remains unchanged when the function is applied. This concept is crucial when working with equations in a CPO, as we look for the smallest subset that resolves an equation. In our exercises, the task is to find such least fixpoints for a given function. The method often involves starting from the least element, typically \( \perp \), which represents an empty set or the smallest element in that context. Then, through iterative application of an order-preserving map, we converge to the minimal set that solves the equation.
Order-Preserving Map
An order-preserving map, often denoted as \((F)\), is a function between posets (partially ordered sets) where the order of elements is retained. In essence, if \(a \leq b\), then \(F(a) \leq F(b)\). This property is vital in the construction of fixpoints within a CPO because it ensures that the process of iteratively applying the function always maintains the order of elements. In practical terms, when seeking a least fixpoint, using an order-preserving map \(F\) allows us to start from a known point \(\perp\) and apply \(F\) repetitively, thus ensuring we are building up in an organized manner. Each stage \(F^n(\perp)\) results in a more defined subset, bringing us incrementally closer to the final fixpoint.
Iterative Method
The iterative method is a systematic approach frequently used to find the least fixpoint in a complete partial order. It involves starting from the least element \(\perp\), generally the empty set, and applying a function repeatedly. This function should be an order-preserving map. The key advantage of this method is its structured process, where with each application, information is maintained and expanded systematically. We iterate until the process stabilizes or converges, meaning additional applications of the function cease to produce different results. This convergence is what confirms the attainment of the least fixpoint. By iteration, we can manage complex calculations in steps, building solutions incrementally and verifying the correctness at each stage.
Subgroup Generation
In the context of algebra, particularly group theory, subgroup generation refers to generating a new subgroup from a subset of group elements. In the given exercise, finding the least fixpoint involves iterating through a function to achieve this generation. For example, given a group \(X\) and a subset \(T\), we use a function like \(F(S) = T \cup S \cup S \cdot S\) to repeatedly combine and expand elements, ensuring that closure properties of groups such as associativity and the ability to form inverses are respected. By using the iterative process with this function, we eventually capture all necessary products and powers formed from the initial subset, thereby generating the smallest subgroup containing all these elements while maintaining group structure.

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Most popular questions from this chapter

For each partial map \(\pi\) on \(\mathbb{N}\) define another such map by \(F(\pi)=\pi\) where $$ \pi(k)=\left\\{\begin{array}{ll} 1 & \text { if } k=1 \\ \pi(k-1)+k & \text { if } k>1 \end{array}\right. $$ (Here \(\bar{\pi}\) is to have the maximum domain allowed by its specification \((\operatorname{see} 8.17)\), so that \(\operatorname{dom} \bar{\pi}=\\{1\\} \cup\\{k+1 \mid k \in \operatorname{dom} \pi\\} .)\) Let \(\pi\) be a solution in \((\mathbb{N} \multimap \rightarrow \mathbb{N})\) to the fixpoint equation \(F(\pi)=\pi .\) Verify that \(\pi\) is total and that \(\pi(k+1)=\pi(k)+k+1\), for all \(k \in \mathbb{N}\). Deduce, by induction, that \(\pi(k)=\sum_{i=1}^{k} i\), for all \(k \in \mathbb{N}\).

Let \(P\) be a pre-CPO and let \(A_{i} \subseteq P\) for all \(i \in I .\) Suppose that \(A_{i}\) is a directed subset of \(P\) for all \(i \in I\) and that \(\left\\{A_{i}\right\\}_{t \in I}\) is a directed family of subsets of \(P\). Show that both \(\bigcup_{i \in I} A_{i}\) and \(\left\\{\sqcup A_{i}\right\\}_{i \in I}\) are directed subsets of \(P\) and that $$ \bigcup\left(\bigcup_{i \in I} A_{i}\right)=\bigcup_{i \in I}\left(\sqcup A_{i}\right) . $$

Let \(F\) be an order-preserving self-map on a complete lattice \(P\). Prove that \(\mu(F \circ F)=\mu(F)\) (the Square Rule).

(For those who know about ordinals.) Let \(P\) be a \(\mathrm{CPO}\) and \(F: P \rightarrow P\) an order-preserving map. Let $$ \begin{aligned} F^{0}(\perp) &=\perp \\ F^{\beta+1}(\perp) &=F\left(F^{\beta}(\perp)\right) \end{aligned} $$ \(F^{\alpha}(\perp)=\bigcup\left\\{F^{\beta}(\perp) \mid \beta<\alpha\right\\}\) if \(\alpha\) is a limit ordinal. Prove that \(F^{\alpha}(\perp)\) is well defined for all ordinals \(\alpha\) by showing (by transfinite induction) that, if \(\alpha \leqslant \beta\) and \(F^{\beta}(\perp)\) is defined, then \(F^{\alpha}(\perp) \leqslant F^{\beta}(\perp) .\) Argue by contradiction to show (with the aid of Hartogs' Theorem) that, for some ordinal, \(\alpha\), the point \(F^{\alpha}(\perp)\) is a fixpoint of \(F\).

Given \(f: \mathbb{N}_{0} \rightarrow\left(\mathbb{N}_{0}\right)_{\perp}\), let \(\bar{f}\) be defined by $$ \bar{f}(k)=\left\\{\begin{array}{ll} 1 & \text { if } k=0 \\ k f(k-1) & \text { if } k \geqslant 1 \text { and } f(k-1) \neq 1 \\ \perp & \text { otherwise } \end{array}\right. $$ Define \(F:\left(\mathbb{N}_{0} \rightarrow\left(\mathbb{N}_{0}\right)_{\perp}\right) \rightarrow\left(\mathbb{N}_{0} \rightarrow\left(\mathbb{N}_{0}\right)_{\perp}\right)\) by \(F(f)=\bar{f}\). (i) Show that \(F\) is order-preserving: (ii) Prove that \(F\) is continuous. [Hint. Let \(D=\left\\{g_{i}\right\\}_{i \in I}\) be a directed set in \(\left(\mathrm{N}_{0} \rightarrow\left(\mathbb{N}_{0}\right)_{\perp}\right)\), and let \(g:=\bigcup_{i \in I} g_{i} .\) Check that \(\bar{g}(k)=\bigcup_{i \in I} \bar{g}_{i}(k)\) for all \(k \in \mathbb{N}_{0}\) by considering the cases (a) \(k=0\) (b) \(g_{i}(k-1)=\perp\) for all \(i \in I\) (c) \(g_{i}(k-1) \neq \perp\) for some \(\left.i \in I .\right]\) (This exercise establishes the continuity of the map of which the factorial function is the least fixpoint.)
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