Deleting one point of a countably infinite set leaves a set equinumerous with it

For $n\in \mathbb{N}$ put ${\phi }_n:=\{(k,v)\in \mathbb{N}\times \mathbb{N}:(k<n\wedge v=k)\vee (n\le k\wedge v=\sigma (k))\}$, a set by Separation, and read it as a map. It is a function $\mathbb{N}\to \mathbb{N}\setminus \{n\}$. Indeed, for each $k\in \mathbb{N}$ the two clauses are exhaustive and mutually exclusive: if $k<n$ then $n\le k$ fails, since $k<n$ gives $k\le n$ and $k\mathrm{\ne }n$ while $n\le k$ would let antisymmetry give $k=n$; and if $k<n$ fails then $k=n$ or $n<k$ by trichotomy, so $n\le k$ holds by reflexivity or by $n<k$ implying $n\le k$. Hence exactly one value $v$ is paired with $k$, and $v$ is $k\in \mathbb{N}$ or $\sigma (k)\in \mathbb{N}$, so $\mathrm{dom}{\phi }_n=\mathbb{N}$. The values avoid $n$: in the first clause $v=k\mathrm{\ne }n$; in the second clause $v=\sigma (k)\mathrm{\ne }n$, because $\sigma (k)=n\le k$ would give $\sigma (k)\le k$ and hence $k<k$ by discreteness, which is impossible since $k<k$ would require $k\mathrm{\ne }k$.

$X\setminus \{x\}$ is a proper subset of $X$. By [F7] it is a subset of $X$; it differs from $X$ because $x\in X$ while $x\mathrm{\notin }X\setminus \{x\}$, as $x=x$.

No finite set is equinumerous with a proper subset of itself. Let $F$ be finite, let $A\subseteq F$ with $A\mathrm{\ne }F$, and suppose towards a contradiction that $h:F\to A$ is a bijection. Fix $k\in \mathbb{N}$ and a bijection $\psi :F\to k$, possible since $F$ is finite. Then $A\approx \psi [A]$ via the restriction of $\psi$ to $A$ by [F5], and $\psi [A]\approx A\approx F\approx k$ by symmetry and transitivity of $\approx$, so $\psi [A]\subseteq k$ satisfies $k\approx \psi [A]$; claim 5 of [F6] gives $\psi [A]=k$. Then $F\subseteq A$: for $f\in F$ the value $\psi (f)\in k=\psi [A]$ equals $\psi (a)$ for some $a\in A$, so $f=a$ by injectivity of $\psi$. With $A\subseteq F$ this gives $A=F$, contradicting $A\mathrm{\ne }F$.

For $X=\mathbb{N}$ and $x=0$ the witness is the successor map: $\sigma$ is injective and never zero by [F2], so it is a function $\mathbb{N}\to \mathbb{N}\setminus \{0\}$, and it is onto $\mathbb{N}\setminus \{0\}$ because every $n\mathrm{\ne }0$ is a successor by [F2]. Hence $\sigma$ is a bijection $\mathbb{N}\to \mathbb{N}\setminus \{0\}$.

Since $X$ is countably infinite, [F1] supplies a bijection $e:\mathbb{N}\to X$; let $n\in \mathbb{N}$ with $e(n)=x$, which exists because $e$ is onto. Then $e[\mathbb{N}\setminus \{n\}]=X\setminus \{x\}$: if $t=e(k)$ with $k\mathrm{\ne }n$ then $t\mathrm{\ne }x$, since $t=x=e(n)$ would force $k=n$ by injectivity of $e$; conversely, if $t\in X$ and $t\mathrm{\ne }x$, then $t=e(k)$ for some $k$ by surjectivity and $k\mathrm{\ne }n$ since $e(n)=x\mathrm{\ne }t$.

${\phi }_n$ is injective. Let ${\phi }_n(k)={\phi }_n(l)$ and split each of $k,l$ by trichotomy; the alternatives $k=n$ and $n<k$ both assign the value $\sigma (k)$, and $n\le k$ in both. If $k<n$ and $l<n$ the values are $k$ and $l$, so $k=l$. If $n\le k$ and $n\le l$ the values are $\sigma (k)$ and $\sigma (l)$, so $k=l$ by injectivity of $\sigma$. In a mixed case, say $k<n$ and $n\le l$, the values are $k$ and $\sigma (l)$ with $l=n$ or $n<l$: if $l=n$ then $k=\sigma (n)\le n$ because $k<n$ gives $k\le n$, so $\sigma (n)\le n$ and discreteness gives $n<n$, impossible; if $n<l$ then $k=\sigma (l)\le n$, so discreteness gives $l<n$, contradicting $n<l$ by trichotomy. The case $n\le k$ with $l<n$ is symmetric. Hence $k=l$ in every case.

The restriction $e_0$ of $e$ to $\mathbb{N}\setminus \{n\}$, read as a map into $X\setminus \{x\}$, is a bijection $\mathbb{N}\setminus \{n\}\to X\setminus \{x\}$: it is injective, being a restriction of the injection $e$, and it is onto, since its image is $e[\mathbb{N}\setminus \{n\}]=X\setminus \{x\}$ by step 1.5.

${\phi }_n$ is surjective onto $\mathbb{N}\setminus \{n\}$. Let $m\in \mathbb{N}$ with $m\mathrm{\ne }n$. If $m<n$ then ${\phi }_n(m)=m$; if $n<m$ then $n\le m$ with $n\mathrm{\ne }m$, so $m=n+j$ for some $j\mathrm{\ne }0$, $j=\sigma (i)$ by [F2], and $m=n+\sigma (i)=\sigma (n+i)$ by [F3]. Then $k:=n+i$ satisfies $n\le k$ (witnessed by $i$) and ${\phi }_n(k)=\sigma (k)=m$. So ${\phi }_n$ is onto $\mathbb{N}\setminus \{n\}$, and with step 2.1 it is a bijection $\mathbb{N}\to \mathbb{N}\setminus \{n\}$.

The composite $e_0\circ {\phi }_n\circ e^{-1}:X\to X\setminus \{x\}$ is a bijection: $e^{-1}$ is a bijection $X\to \mathbb{N}$ by [F5], ${\phi }_n$ is a bijection $\mathbb{N}\to \mathbb{N}\setminus \{n\}$ by step 3.1, and $e_0$ is a bijection $\mathbb{N}\setminus \{n\}\to X\setminus \{x\}$ by step 2.2, so by [F5] the composite is a bijection. Hence $X\approx X\setminus \{x\}$.

Therefore $X\approx X\setminus \{x\}$ by step 4.1, the set $X\setminus \{x\}$ is a proper subset of $X$ by step 1.2, the case $X=\mathbb{N}$, $x=0$ is the successor bijection of step 1.4, and no finite set is equinumerous with a proper subset of itself by step 1.3. Every map used is defined by an explicit rule or is the unique inverse of a bijection, so no choice principle is used.