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Chapter 8: Problem 5

Let \(P_{1}\) and \(P_{2}\) be CPOs. Let \(D \subseteq P_{1} \times P_{2}\) be a directed set. Define \(D_{1}\) and \(D_{2}\) by $$ \begin{aligned} &D_{1}:=\left\\{x_{1} \in P_{1} \mid\left(\exists x_{2} \in P_{2}\right)\left(x_{1}, x_{2}\right) \in D\right\\} \\ &D_{2}:=\left\\{x_{2} \in P_{2} \mid\left(\exists x_{1} \in P_{1}\right)\left(x_{1}, x_{2}\right) \in D\right\\} \end{aligned} $$ Show that \(D_{1}\) and \(D_{2}\) are directed and that \(\bigcup D=\left(\sqcup D_{1}, \bigcup D_{2}\right)\). Deduce that \(P_{1} \times P_{2}\) is a CPO.

Short Answer

Expert verified
\(D_1\) and \(D_2\) are directed, and \(\bigcup D = (\sqcup D_1, \sqcup D_2)\), so \(P_1 \times P_2\) is a CPO.

Step by step solution

01

Show that D鈧� is directed

To prove that \(D_1\) is directed, we need to show that for any \(x_1, y_1 \in D_1\), there exists \(z_1 \in D_1\) such that \(x_1 \leq z_1\) and \(y_1 \leq z_1\). Given any \(x_1, y_1 \in D_1\), there exist \(x_2, y_2 \in P_2\) such that \((x_1, x_2) \in D\) and \((y_1, y_2) \in D\). Since \(D\) is directed, there exists \((z_1, z_2) \in D\) such that \((x_1, x_2) \leq (z_1, z_2)\) and \((y_1, y_2) \leq (z_1, z_2)\). Thus, \(x_1 \leq z_1\) and \(y_1 \leq z_1\), showing that \(D_1\) is directed.
02

Show that D鈧� is directed

Similarly, for \(D_2\), assume we have \(x_2, y_2 \in D_2\). There exist \(x_1, y_1 \in P_1\) such that \((x_1, x_2) \in D\) and \((y_1, y_2) \in D\). Since \(D\) is directed, there is some pair \((z_1, z_2) \in D\) such that \((x_1, x_2) \leq (z_1, z_2)\) and \((y_1, y_2) \leq (z_1, z_2)\). Therefore, \(x_2 \leq z_2\) and \(y_2 \leq z_2\) which shows that \(D_2\) is directed.
03

Establish the union of D in terms of suprema

To show that \(\bigcup D = (\sqcup D_1, \sqcup D_2)\), we need to demonstrate that the supremum of \(D\) in \(P_1 \times P_2\) is obtained by taking the supremum of \(D_1\) in \(P_1\) and \(D_2\) in \(P_2\). Since \(D\) is directed, it will have a supremum \((\sqcup D_1, \sqcup D_2)\), where \(\sqcup D_1\) is the least upper bound of \(D_1\) and \(\sqcup D_2\) is the least upper bound of \(D_2\). It follows that \((\sqcup D_1, \sqcup D_2)\) is indeed the supremum of \(D\).
04

Conclude that P鈧� 脳 P鈧� is a CPO

A poset is a CPO if every directed subset has a supremum. Since we've shown that any directed subset \(D\) of \(P_1 \times P_2\) has a supremum \((\sqcup D_1, \sqcup D_2)\), it follows that \(P_1 \times P_2\) is a CPO, as it retains the completeness property within its poset structure.

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Key Concepts

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

Directed set
In the context of posets, a directed set is a subset where every pair of elements has an "upper bound" within the same set. This captures the idea of elements "heading" in a similar direction without contradiction. In simpler terms, take any two elements in the directed set; you can always find another element in the same set that's larger than or equal to both of them.
For example, consider a directed set within a poset. If you have two numbers, say 3 and 7, in this set, you should be able to find another number, perhaps 10, that's greater than or equal to 3 and 7.
Understanding directed sets is crucial for working with complete partial orders (CPOs) because they enable the definition of elements that can potentially reach a supremum or least upper bound.
Supremum
The supremum of a set refers to the "least upper bound." This means it is the smallest possible number that is greater than or equal to every element in the set. It鈥檚 not necessarily an element of the set but is often called the 'least' because it鈥檚 the lowest among all upper bounds.
Suppose you have the set of numbers {1, 2, 3}. Here, the supremum is 3, as it is the maximum number within the set. However, in the case of an open interval like (0,1), the supremum is 1, even though 1 is not included in the interval. This makes supremum a very flexible concept when describing upper bounds in more abstract sets.
In CPOs, knowing the supremum helps to determine the structure and bounds within the poset, ensuring every directed set has a clear upper limit.
Least upper bound
The least upper bound is alternatively known as the supremum but from a more intuitive angle. It is the smallest value that is still an upper bound for the set. Usefully, it's the "ceiling" that can be reached by any elements from within the set, given the constraints of the set.
If you have a directed set, say {5, 10, 15}, your least upper bound is 15. No number smaller than 15 can serve as a global upper bound without violating the "upper" property of lesser elements in the set.
  • It helps encapsulate all elements without leaving room for any to exceed the limit.
  • Unlike maximums, least upper bounds work even if not part of the set.
  • Vital in mathematical operations across diverse sets, ensuring completeness within CPOs.
Poset
A poset stands for "Partially Ordered Set," which is a foundational structure in order theory. It consists of a set of elements with a binary relation that describes the order between them.
For a set to be considered a poset, it must satisfy three conditions:
  • Reflexivity: Every element is related to itself, e.g., if you're considering numbers, 3 is related to 3.
  • Antisymmetry: If one element is related to another, and vice versa, then they are identical. For instance, if 4 鈮� 5 and 5 鈮� 4 both hold, then 4 and 5 must be the same, which doesn't happen, keeping the set antisymmetric.
  • Transitivity: If an element is less than or equal to a second, and this second one is less than or equal to a third, the first is also less than or equal to the third.

A poset becomes a CPO when every directed subset has a supremum, giving it a complete, closed structure which aids in various computations and logical deductions.

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Most popular questions from this chapter

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

Let \(L\) be a complete lattice and define \(F: \wp(L) \rightarrow \wp(L)\) by \(A \mapsto \downarrow \bigvee A\). Show that \(F\) is order-preserving and that \(f x(F) \cong L\) (This is the long proof that every complete lattice is, up to isomorphism, a lattice of fixpoints. Find the short proof.)

Let \(P\) be a complete lattice and \(F: P \rightarrow P\) be order-preserving. Let \(X\) be a subset of \(f x(F) .\) Define $$ Y=\\{y \in P \mid(\forall x \in X) x \leqslant F(y) \leqslant y\\} $$ and let \(\alpha=\bigwedge_{P} Y .\) Prove that \(\alpha=V_{\mathrm{fix}(F)} X .\) Deduce that \(\operatorname{fix}(F)\) is a complete lattice.

Let \(P\) be an ordered set and \(Q\) a complete lattice. Given a map \(f: P \rightarrow Q\), define \(\bar{f}: P \rightarrow Q\) by $$ \bar{f}(x)=\bigvee\\{f(y) \mid y \leqslant x\\} $$ Show that \(\bar{f}\) is order-preserving and that \(f=\bar{f}\) if and only if \(f\) is order-preserving. Show further that \(F\) defined by \(F: f \mapsto \bar{f}\) is an order-preserving map from \(Q^{P}\) to \(Q^{(P)} \subseteq Q^{P}\) whose set of fixpoints is exactly \(Q^{(P\rangle}\).

Let \(P\) be a CPO. Let \(\mu:[P \rightarrow P] \rightarrow P\) be the fixpoint operator which assigns to each continuous map \(f: P \rightarrow P\) its least fixpoint \(\mu(f)\) and, for each \(n \in \mathbb{N}\), let \(\mu_{n}\) be defined by \(\mu_{n}(f)=f^{n}(\perp)\) (i) Show, by induction, that \(\mu_{n}\) is continuous for each \(n\). (ii) Let \(Q=([P \rightarrow P] \rightarrow P)\), the CPO of all maps from \([P \rightarrow P]\) to \(P .\) Show that \(\left\\{\mu_{n}\right\\}_{n \geqslant 1}\) is a chain in \(Q\). (iii) Show that \(\bigcup_{n \geqslant 1} \mu_{n}=\mu\) in \(Q .\) Deduce that \(\mu\) is continuous. (This strengthens the monotonicity assertion in Exercise \(8.25(\mathrm{i}) .\) )
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