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Let \(\mathfrak{A}=\langle A, \leq\rangle\) be a poset. Show that \(\operatorname{Diag}^{+}(\mathfrak{A}) \cup\\{\bar{a} \neq \bar{b} \mid a \neq b, a, b \in\) \(|\mathfrak{A}|\\} \cup\\{\forall x y(x \leq y \vee y \leq x)\\}\) has a model. (Hint: use compactness). Conclude that every poset can be linearly ordered by an extension of its ordering.

Short Answer

Expert verified
Any poset can be linearly ordered by extending its ordering using compactness.

Step by step solution

01

Understanding the Problem Statement

We have a poset \(\mathfrak{A}=\langle A, \leq\rangle\) and we need to show that a certain logical set \(\operatorname{Diag}^{+}(\mathfrak{A}) \cup\{\bar{a} eq \bar{b} \mid a eq b, a, b \in |\mathfrak{A}|\} \cup\{\forall x y(x \leq y \vee y \leq x)\}\) has a model. This set aims to describe a linear order consistent with the original partial order.
02

Break Down the Logical Set

1. \(\operatorname{Diag}^{+}(\mathfrak{A})\) includes all the atomic formulas and their negations that hold in \(\mathfrak{A}\).2. \(\{\bar{a} eq \bar{b} \mid a eq b, a, b \in |\mathfrak{A}|\}\) ensures distinctness among different elements.3. \(\{\forall x y(x \leq y \vee y \leq x)\}\) adds the condition that any two elements \(x\) and \(y\) are comparable, making the order linear.
03

Apply Compactness Theorem

The Compactness Theorem in logic implies that if every finite subset of a set of sentences has a model, then the whole set has a model. Consider finite subsets of our extended logical set, including \(\forall xy(x \leq y \vee y \leq x)\), and show that each subset has a model. Thus, by compactness, the entire set has a model.
04

Conclusion from Compactness

Since every finite subset of sentences has a model, by compactness, the entire set has a model. This model is a linear order because it satisfies \(\forall xy(x \leq y \vee y \leq x)\). Thus, every poset can be extended to a linear order, ensuring the ordering properties are preserved.

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Key Concepts

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

Compactness Theorem
The Compactness Theorem is a fundamental result in mathematical logic, particularly in model theory. It states that for any set of logical sentences, if every finite subset of those sentences has a model, then the entire set also has a model.

This theorem is powerful because it allows us to infer the existence of a model for an infinite set of sentences, simply by checking finite subsets. This ability is useful in various areas of logic and structure, as it bridges the gap between finite and infinite cases.

In the context of partial orders, the compactness theorem helps us to construct a linear extension. By ensuring that every finite subset of a logical set describing a poset has a model, we can conclude that the entire set has a model. Thus, we can extend the original partial order into a total order, or linearization.
Partial Order
A partial order is a binary relation over a set that satisfies three properties: reflexivity, antisymmetry, and transitivity. This means that for a set \(A\) with a partial order \( \leq \), the following must hold:

  • Reflexivity: For any element \(a\) in \(A\), \(a \leq a\).
  • Antisymmetry: For any elements \(a\) and \(b\) in \(A\), if \(a \leq b\) and \(b \leq a\), then \(a = b\).
  • Transitivity: For any elements \(a\), \(b\), and \(c\) in \(A\), if \(a \leq b\) and \(b \leq c\), then \(a \leq c\).
Partial orders do not require comparability between all pairs of elements, which differentiates them from total orders. However, using linear extensions, we can create a total order from a partial one. This is where poset linearization becomes vital, melding partly ordered sets into a fully linear or total order.
Logic and Structure
In logic, structure refers to the mathematical constructs used to interpret formal languages. These structures consist of a domain along with functions and relations over that domain.

Within posets, the logic describes how elements relate to each other. When extending a partial order to a linear order, additional logical structures make every pair of elements comparable. This transformation is explained with logical sentences that enforce linear behavior, ensuring no conflicts in the extended relations.

The application of logic in defining structures like posets provides a rigorous interpretation of relationships within the set. In our exercise, understanding these relationships ensures that our poset is consciously converted into a logically sound linear or total order.
Model Theory
Model Theory studies structures, or models, that satisfy formal languages' given sentences or theories. It connects logic and mathematics by exploring the properties of models that express logical truths.

Within this context, a 'model' is a mathematical structure that satisfies the sentences describing a poset. Our task involved extending a structure described by certain conditions into a linear model.

By applying the model theory concepts, we know that if a poset and its linear extension conditions form sentences within a logical language, and if every finite subset of these sentences, as per the Compactness Theorem, has a model, then the entire set does. This helps us form the necessary model satisfying all conditions of a linear extension of the initial partial order, thus verifying the completeness and consistency of the logical conversions undertaken.

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Most popular questions from this chapter

Let \(\mathfrak{A} \subseteq \mathfrak{B} . \varphi\) is called universal (existential) if \(\varphi\) is prenex with only universal (existential) quantifiers. (i) Show that for universal sentences \(\varphi \mathfrak{B} \models \varphi \Rightarrow \mathfrak{A} \models \varphi\). (ii) Show that for existential sentences \(\varphi \mathfrak{A} \models \varphi \Rightarrow \mathfrak{B} \models \varphi\). (Application: a substructure of a group is a group. This is one reason to use the similarity type \(\langle-; 2,1 ; 1\rangle\) for groups, instead of \(\langle-; 2 ; 0\rangle\), or \(\langle-; 2 ; 1\rangle\), as some authors do).

Let \(X \subseteq|\mathfrak{A}| .\) Define \(X_{0}=X \cup C\) where \(C\) is the set of constants of \(\mathfrak{A}, X_{n+1}=X_{n} \cup\left\\{f\left(a_{1}, \ldots, a_{m}\right) \mid f\right.\) in \(\left.\mathfrak{A}, a_{1}, \ldots, a_{m} \in X_{n}\right\\}, X_{\omega}=\) \(\bigcup\left\\{X_{n} \mid n \in \mathbb{N}\right\\}\) Show that \(\mathfrak{B}=\left\langle X_{\omega}, R_{1} \cap X_{\omega}^{r_{1}}, \ldots, R_{2} \cap X_{\omega}^{r_{1}}, f_{1}\left|X_{\omega}^{a_{1}}, \ldots, f_{m}\right| X_{\omega}^{a_{m}},\left\\{c_{1} \mid i \in\right.\right.\) \(I\\}\rangle\) is a substructure of \(\mathfrak{A}\). We say that \(\mathfrak{B}\) is the substructure generated by \(X\). Show that \(\mathfrak{B}\) is the smallest substructure of \(\mathfrak{A}\) containing \(X ; \mathfrak{B}\) can also be characterized as the intersection of all substructures containing \(X\).

Consider the class \(\mathcal{K}\) of all structures of type \(\langle 1 ;-; 0\rangle\) with a denumerable unary relation. Show that any \(\mathfrak{A}\) and \(\mathfrak{B}\) in \(\mathcal{K}\) of the same cardinality \(\kappa>\aleph_{0}\) are isomorphic. Show that \(T=T h(\mathcal{K})\) is not \(\kappa\)-categorical for any \(\kappa \geq N_{0}\).

Determine the Skolem forms of (a) \(\forall y \exists x\left(2 x^{2}+y x-1=0\right)\), (b) \(\forall \varepsilon \exists \delta(\varepsilon>0 \rightarrow(\delta>0 \wedge \forall x(|x-\bar{a}|<\delta \rightarrow|f(x)-f(\bar{a})|<\varepsilon)\), (c) \(\forall x \exists y(x=f(y))\), (d) \(\forall x y(x

Consider a sequence of theories \(T_{i}\) such that \(T_{i} \neq T_{i+1}\) and \(T_{i} \subseteq T_{i+1}\). Show that \(\cup\left\\{T_{i} \mid i \in \mathcal{N}\right\\}\) is not finitely axiomatisable.

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