Isomorphism between Ordered Sets

Question

The proof in Proof Wiki states that the proofs for "empty mapping is mapping" and "mappings between sets of same cardinality are equivalent" imply that $\emptyset\in A$.

Why is that? Can somebody give me an explanation for this, please?

Answer 1

$A$ is the set of all $n\in\Bbb N$ such that if $S$ is any set of cardinality $n$, and $\preceq$ is any total order on $S$, then there is exactly one isomorphism from $\langle S,\preceq\rangle$ to $\langle\Bbb N_n,\le\rangle$. We want to show that $\varnothing\in A$.

The only set $S$ of cardinality $0$ is $\varnothing$, and the result Empty Mapping Is Mapping says that $\varnothing$ is a mapping from $\varnothing$ to any set. In particular, it’s a mapping from $\varnothing$ to $\Bbb N_0$. In the notation used in your source $\Bbb N_0$ is the set of natural numbers less than $0$, so $\Bbb N_0=\varnothing$: $\varnothing$ is a mapping from $\varnothing$ to $\varnothing$.

This mapping is surjective: $\forall x\in\Bbb N_0\,\exists y\in\varnothing(\langle y,x\rangle\in\varnothing)$ is vacuously true simply because there is no $x\in\Bbb N_0$. The result Equivalence of Mappings between Sets of Same Cardinality then says that this mapping is bijective, since its domain $\varnothing$ and range $\Bbb N_0$ are clearly of the same cardinality: they’re even the same set.

We now have a bijection from $\varnothing$ to $\Bbb N_0$, namely, the mapping $\varnothing$. It is vacuously order-preserving: the only total order on $\varnothing$ is $\varnothing$, which is also the only total order on $\Bbb N_0$, and since $\varnothing$ has no members, there’s simply nothing to check. Thus, the map $\varnothing$ is an order-isomorphism from $\langle\varnothing,\varnothing\rangle$ to $\langle\Bbb N_0,\le\rangle$ (where in this case $\le$ turns out to be just $\varnothing$).

Finally, any order-isomorphism from $\langle\varnothing,\varnothing\rangle$ to $\langle\Bbb N_0,\le\rangle$ is a subset of $\varnothing\times\Bbb N_0$, and $\varnothing$ is the only subset of $\varnothing\times\Bbb N_0$, so $\varnothing$ must be the only order-isomorphism from $\langle\varnothing,\varnothing\rangle$ to $\langle\Bbb N_0,\le\rangle$. This shows that the map $\varnothing$ satisfies the requirements for membership in $A$.