Show that the least upper bound axiom also holds in ℤ (i.e., each nonempty subset of ℤ with an u

step 1 Here in this question it is asked to make a statement expressing a property analogous to the lea

Show that the least upper bound axiom also holds in ℤ (i.e.,...

Show that the least upper bound axiom also holds in $\mathbb{Z}$ (i.e., each nonempty subset of $\mathbb{Z}$ with an upper bound in $\mathbb{Z}$ has a least upper bound in $\mathbb{Z}$ ), but that it fails to hold in $\mathbb{Q}$.

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Key Concepts

Least Upper Bound Axiom

This axiom, also known as the supremum property, states that every nonempty subset of an ordered set that has an upper bound must have a least upper bound within that set. It is a fundamental property in analysis used to define completeness of an ordered structure.

Upper Bound

An upper bound of a subset in an ordered set is an element that is greater than or equal to every element in that subset. Identifying upper bounds allows us to consider the 'least' among them, leading to the concept of the least upper bound.

Completeness

Completeness in an ordered set refers to the property that every subset that is bounded above has a least upper bound in the set. This concept distinguishes complete ordered sets from those that are incomplete, where some subsets do not have a least upper bound within the set itself.

Discrete versus Dense Orders

This concept pertains to the structural differences in ordered sets. In discrete ordered sets, elements have immediate successors or predecessors, which often makes the existence of a least upper bound more straightforward. In contrast, in dense ordered sets, between any two elements there are infinitely many other elements, which can result in subsets lacking a least upper bound within the set, thereby exhibiting incompleteness.

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