Non-standard model of arithmatic

Theorem

Let $T=Th(\mathfrak{N})$ be the set of all sentences true in the standard model of arithmetic, $\mathfrak{N}=⟨\mathbb{N},+,\cdot ,<,0,1⟩.$ Then there are models $\mathfrak{A}\models T$ containing an element $a$ such that $1<a,1+1<a,1+1+1<a,\dots$ are all true in $\mathfrak{A}$.

Remark

Because $a$ cannot be reached from $0$ in a finite number of steps, it is an infinite element. Any model containing such an element is “non-standard” because its domain is strictly larger and structured differently than $\mathbb{N}$.

Dense Linear Orderings

A linear ordering $\mathfrak{L}=⟨L,<⟩$ is dense if $\forall x\forall y\left((x<y)\to \exists z(x<z<y)\right).$ $\mathfrak{L}$ has no endpoints if $\forall x\left(\exists y(y<x)\wedge \exists z(x<z)\right).$

End Extension

We say that $\mathfrak{A}$ is an end extension of $\mathfrak{N}$, denoted as $\mathfrak{N}{\subseteq }_{\text{end}}\mathfrak{A}$, if and only if the following two conditions hold:

1. $\mathfrak{N}$ is a submodel of $\mathfrak{A}$, meaning $|\mathfrak{N}|\subseteq |\mathfrak{A}|$, and the interpretations of all constants, functions, and relations in $\mathfrak{A}$ agree with those in $\mathfrak{N}$ when restricted to $|\mathfrak{N}|$. $\mathfrak{N}\subseteq \mathfrak{A}$
2. Any element in $\mathfrak{A}$ that is strictly less than an element in $\mathfrak{N}$ must already belong to $\mathfrak{N}$. In other words, $\mathfrak{N}$ forms an initial segment of $\mathfrak{A}$ with respect to the ordering $<$. $\forall x\in |\mathfrak{A}|((\exists y\in |\mathfrak{N}|\text{ such that }x{<}^{\mathfrak{A}}y)⟹x\in |\mathfrak{N}|)$

Alternatively, this second condition can be stated in its contrapositive form, which directly describes the “new” elements: All new elements added to the extension $\mathfrak{A}$ are strictly greater than every element in the original model $\mathfrak{N}$.

$\forall a\in (|\mathfrak{A}|\setminus |\mathfrak{N}|),\forall m\in |\mathfrak{N}|⟹m{<}^{\mathfrak{A}}a$

Isomorphic Orderings

Two (linear) orderings $\mathfrak{A}=⟨A,{<}_A⟩$ and $\mathfrak{B}=⟨B,{<}_B⟩$ are isomorphic, written $\mathfrak{A}\cong \mathfrak{B}$, if there exists a bijection $f:A\to B$ such that $a{<}_A{a}^{\prime }$ iff $f(a){<}_{B}f(a^{\prime })$.

Theorem

Suppose $\mathfrak{A}=⟨A,{<}_A⟩$ and $\mathfrak{B}=⟨B,{<}_B⟩$ are countable dense linear orders without end points. Then they are isomorphic.