Question

Consider the structure \(\mathfrak{A}=\langle\mathbb{R},\langle, f\rangle\), where \(<\) is the natural order, and where \(f\) is a unary function. Let \(L\) be the corresponding language. Show that there is no sentence \(\sigma\) in \(L\) such that \(\mathfrak{A} \models \sigma \Leftrightarrow f(r)>0\) for all \(r \in R\). (hint: consider isomorphisms \(x \mapsto x+k\) ).

Short answer

No such sentence \(\sigma\) exists due to potential contradictions from isomorphic transformations.

Step-by-step solution

Understanding the Structure

The structure \(\mathfrak{A}=\langle\mathbb{R},<, f\rangle\) consists of the set of real numbers \(\mathbb{R}\), equipped with the standard order, and a unary function \(f\). The language \(L\) contains symbols for the order \(<\) and the function \(f\). We aim to show there does not exist a sentence \(\sigma\) such that \(\mathfrak{A} \models \sigma \Leftrightarrow f(r)>0\) for all \(r \in \mathbb{R}\).

Analyzing the Isomorphism

Consider an isomorphism of \(\mathbb{R}\) of the form \(x \mapsto x+k\), where \(k \in \mathbb{R}\). This transformation preserves the order properties of the reals because if \(x < y\), then \(x+k < y+k\). The main focus is to understand how \(f\) behaves under this transformation.

Effect on the Function

For this setup to alter the function \(f\), if \(f(r) > 0\) for all \(r\), then \( (f(r+k) > 0) \) must hold under all considered isomorphisms. If the function \(f\) is constant and positive everywhere, it holds. However, if \(f\) is not the constant function, choosing a parameter \(k\) such that \(f(r + k) \leq 0\) for some \(r\) would provide a contradiction.

Demonstrating Non-existence of \(\sigma\)

Since we can always find a shift \(k\) that potentially alters \(f\) so it does not remain positive for every real number, no sentence \(\sigma\) in \(L\) can uniformly express that \(f(r) > 0\) for all \(r \in \mathbb{R}\) across all isomorphic structures. This contrast evidences no such sentence \(\sigma\) exists because \(\sigma\) would have to hold identically in all these isomorphic structures.

Key concepts

Structure Theory

In mathematics, structure theory relates to investigating objects by examining their internal features and relations. For this particular exercise, we focus on the structure \( \mathfrak{A} = \langle \mathbb{R}, <, f \rangle \). Here, the structure includes the set of real numbers \( \mathbb{R} \) with the standard order relation \(<\) and a unary function \(f\). Structure theory helps us understand how different parts of this mathematical setup interact.

• The real numbers are ordered naturally, meaning smaller numbers come before larger ones.
• A unary function \(f\) is a function with one input that maps to an output number.
• The related language \(L\) defines symbols and syntax to express these relationships.

Understanding these structural elements is crucial to analyzing whether a particular property can hold in a given system.

Real Numbers

The real numbers \( \mathbb{R} \) are an essential part of the structure we are examining. These numbers form a complete, ordered field, meaning you can always find a number between two others, and they follow a specific order structure.
Key features of real numbers:

• They include all the rational and irrational numbers.
• Real numbers can be arranged in an increasing order.
• They are central to many branches of mathematics, especially calculus and analysis.

In our exercise, the order relation \(<\) reflects this natural ordering, which helps us analyze how transformations affect the function \(f\) within this ordered field.

Isomorphism

In mathematical logic, an isomorphism implies a structure-preserving mapping between two mathematical structures. Here, we consider an isomorphism of the real numbers specified by the transformation \(x \mapsto x+k\), where \(k\) is any real number.

• This shift keeps the order intact: if \(x < y\), then \(x + k < y + k\).
• It tests how the function \(f\) behaves when the input values are shifted by \(k\).
• The main challenge is to see whether \(f\) still holds the property \(f(r + k) > 0\) for all shifts.

If \(f\) fails to be strictly positive for all real numbers after any shift, then no single sentence \( \sigma \) in \( L \) can universally express that \(f(r) > 0\) for every \(r\) in \( \mathbb{R} \). This understanding is crucial as it reveals the limitations of expressions in the language \(L\) when dealing with isomorphic transformations.

Order Relations

An order relation is a way to arrange the elements of a set in a sequence. In \(\mathbb{R}\), the order \(<\) signifies that any number is either less than, equal to, or greater than any other number.

• The order property is transitive: if \(a < b\) and \(b < c\), then \(a < c\).
• It is anti-symmetric: \(a < b\) and \(b < a\) cannot both be true.
• It is total: any two numbers \(a\) and \(b\) will satisfy one and only one of the relations \(a < b, a = b,\) or \(a > b\).

In our current structure, this order plays a pivotal role in defining whether the function \(f\)'s positivity can be uniformly expressed for all transformations of \(\mathbb{R}\). Changing the order through isomorphic transformations tests the uniformity and consistency of properties across the entire field. Such analysis is critical when we explore the existence of universal sentences like \( \sigma \) in language \( L \).