The one-point extension of a well-order, and $\omega$ with a new top element

The relation $\prec$ is irreflexive: if $x\prec x$ with $x\in V$ then $x<x$, impossible in the total order $V$; if $x=\mathrm{\infty }$ then the definition would need $\mathrm{\infty }\in V$, which is excluded.

Trichotomy: for distinct $x,y\in W$ either both lie in $V$, and totality of $<$ decides between $x<y$ and $y<x$, or one of them is $\mathrm{\infty }$, and then $x\prec \mathrm{\infty }$ when $x\in V$ while $y\prec \mathrm{\infty }$ when $y\in V$.

Transitivity: if $x\prec y$ and $y\prec z$, then either $z=\mathrm{\infty }$, in which case $x\in V$ and $x\prec z$ by the second clause, or $z\in V$; in the second case $y\in V$ and the first clause gives $x<y<z$, so $x<z$ and $x\prec z$.

$W_{<\mathrm{\infty }}=V$, and the identity map is an order isomorphism of $(V,<)$ onto $(W_{<\mathrm{\infty }},\prec )$: a member $x$ of $W$ satisfies $x\prec \mathrm{\infty }$ exactly when $x\in V$, by the second clause of the definition; and for $x,y\in V$ one has $x<y$ if and only if $x\prec y$, so the identity is a bijection onto $W_{<\mathrm{\infty }}$ preserving and reflecting the order.

Every nonempty $S\subseteq W$ has a $\prec$-least element: if $S$ meets $V$, then $S\cap V$ is a nonempty subset of the well-order $V$ with least element $\mathrm{\ell }$, and $\mathrm{\ell }$ is $\prec$-least in $S$, since a member of $S\cap V$ is either $\mathrm{\ell }$ itself or lies strictly above $\mathrm{\ell }$ in $V$, hence in $\prec$, while $\mathrm{\infty }$ satisfies $\mathrm{\ell }\prec \mathrm{\infty }$; if $S$ meets $V$ in nothing, then $S=\{\mathrm{\infty }\}$ and $\mathrm{\infty }$ is least.

For $V=\omega$ the extension is equinumerous with $\omega$: the map $g$ with $g(\mathrm{\infty })=0$ and $g(n)=\sigma (n)$ for $n\in \omega$ is a bijection $W\to \omega$. It is injective, since $\sigma$ is injective and $\sigma (n)\mathrm{\ne }0$ for every $n$, so the values $\sigma (n)$ are nonzero and distinct and differ from $0=g(\mathrm{\infty })$; and it is surjective, since $0=g(\mathrm{\infty })$ while every $m\mathrm{\ne }0$ in $\omega$ equals $\sigma (m^{\mathrm{\prime }})=g(m^{\mathrm{\prime }})$ for some $m^{\mathrm{\prime }}$ by [F9].

When $V=\omega$: $\mathrm{\infty }$ is a greatest element of $W$, because $x\prec \mathrm{\infty }$ for every $x\in V$; and $\omega$ has no greatest element, because $n<\sigma (n)\in \omega$ for every $n\in \omega$.

For $V=\omega$ and $n\in \omega$ the predecessors of $n$ in $W$ are exactly the naturals below $n$: a member $y$ of $W$ satisfies $y\prec n$ if and only if $y\in \omega$ and $y<n$, because $y=\mathrm{\infty }$ satisfies neither disjunct of the definition and a natural $y$ satisfies the first exactly when $y<n$; hence $W_{<n}=\{m\in \omega :m<n\}=n$.

The collapse computes the finite levels: for $V=\omega$ one has $F(n)=n$ for every $n\in \omega$, where $F(y)=\{F(z):z\prec y\}$ is the collapsing map of [F5]. Otherwise the set of $n$ with $F(n)\mathrm{\ne }n$ is nonempty and has a least element $n_0$; then $F(m)=m$ for every $m<n_0$, and by step 1.8 and [F5] we get $F(n_0)=\{F(z):z\prec n_0\}=\{F(m):m<n_0\}=\{m:m<n_0\}=n_0$ by [F10], a contradiction.

Claim (i), the comparability consequence: step 1.4 exhibits an order isomorphism $V\cong W_{<\mathrm{\infty }}$, which is alternative (ii) of [F4] for the pair of well-orders $(V,<)$ and $(W,\prec )$ with witness $\mathrm{\infty }$; by the mutual exclusivity in [F4] the alternative $V\cong W$ fails, and the witness is unique.

Claim (ii), the non-isomorphism: $W$ is not order isomorphic to $(\omega ,<)$. If $f:W\to \omega$ were an order isomorphism, then every $m\in \omega$ is $f(x)$ for some $x\in W$, and every $x\in W$ satisfies $x=\mathrm{\infty }$ or $x\prec \mathrm{\infty }$ because $\mathrm{\infty }$ is greatest by step 1.7, so $m=f(x)\le f(\mathrm{\infty })$; thus $f(\mathrm{\infty })$ would be a greatest element of $\omega$, contradicting step 1.7 since $f(\mathrm{\infty })<\sigma (f(\mathrm{\infty }))$ in $\omega$.

Claim (iii): for $V=\omega$ the predecessors of $\mathrm{\infty }$ are exactly the members of $\omega$ by step 1.4, so $F(\mathrm{\infty })=\{F(z):z\prec \mathrm{\infty }\}=\{F(n):n\in \omega \}=\{n:n\in \omega \}=\omega$ by step 2.1, the last equality because $\{n:n\in \omega \}$ and $\omega$ have the same members; and the image of $F$ is $\mathrm{o}\mathrm{t}(W)$ by [F5], so $\mathrm{o}\mathrm{t}(W)=\{F(y):y\in W\}=\{n:n\in \omega \}\cup \{\omega \}=\omega \cup \{\omega \}={\omega }^{+}$ by [F11]: the collapse is an order isomorphism of $W$ onto the ordinal ${\omega }^{+}$.

Parts (i) to (iii) are established: $\prec$ is irreflexive, trichotomous and transitive by steps 1.1 to 1.3 and every nonempty subset of $W$ has a least element by step 1.5, so $W$ is a well-order with $W_{<\mathrm{\infty }}=V$ by step 1.4; the comparability classification and its unique witness are step 2.2. For $V=\omega$ the extension is equinumerous with $\omega$ by step 1.6 and not order isomorphic to it by step 2.3, and the collapse has image ${\omega }^{+}$ by steps 2.1 and 3.1, so the order type of the extension is ${\omega }^{+}$. Nothing is ever selected: the least elements come from the given well-order of $V$, the bijection in step 1.6 is one explicit formula, and the collapse is unique by [F5], so no choice principle is used.