Is there a definition of finite sets equivalent to the traditional definition (equal cardinality {0, 1, .., n}) in classical mathematics, such that it can be used to constructively prove that a subset and quotient set of a finite set are finite?

It is clear that then this definition cannot be constructively equivalent to the traditional one (since for the traditional one this property is equivalent to the law of the excluded middle).

Or is it proven that no such definition exists?

  • $\begingroup$ Your question isn't very clear: what "property" are your referring to in the second paragraph? If a finite set is defined to be a set in bijection with $\{0, 1, \ldots, n\}$ for some $n$, I don't see why excluded middle is needed to show that subsets of finite sets are finite. $\endgroup$
    – Rob Arthan
    Oct 23 at 20:19
  • $\begingroup$ @RobAtrhan If a finite set is defined to be a set in bijection with $\{0,1,…, n\}$ for some $n$ then the statement "the subset of a finite set is finite" is equivalent to LEM. A class of sets C has the property from paragraph 2 iff every a subset and (after editing the question) every a factor set of a set from C belong to C. $\endgroup$ Oct 23 at 20:35
  • $\begingroup$ Thanks for clarifying the question. Can you give an explanation or a reference for the statement that a subset of a finite set is finite is equivalent to LEM, please. $\endgroup$
    – Rob Arthan
    Oct 23 at 20:38
  • 1
    $\begingroup$ Thanks for that useful reference. $\endgroup$
    – Rob Arthan
    Oct 23 at 20:43
  • 1
    $\begingroup$ @HanulJeon Eschatum Verus didn't say that LEM can be proved from statements of the form $n=m$ but rather that LEM can be proved for such statements. $\endgroup$ Oct 23 at 22:46

There are four definitions which are classically equivalent to being finite. Let $[n]$ be the set $\{0, 1, ..., n - 1\} = \{m \in \mathbb{N} \mid m < n\}$.

Definition 1: A set $S$ is finite if and only if there exists some $n \in \mathbb{N}$ and a bijection $S \cong [n]$.

Definition 2: A set $S$ is subfinite if and only if there is some $n \in \mathbb{N}$ and some injection $S \to [n]$.

Definition 3: A set $S$ is finitely enumerable if and only if there is some $n \in \mathbb{N}$ and some surjection $[n] \to S$.

Definition 4: A set $S$ is sub-finitely enumerable if and only if there is some finitely enumerable $S'$ and some injection $S \to S'$.

Definition 5: A set $S$ is subfinitely enumerable if and only if there is some subfinite set $S'$ and some surjection $S' \to S$.

Note that all these definitions are bijection-invariant. That is, if $A \cong B$ (that is, if there is a bijection $A \to B$), then $A$ satisfies definition $i$ if and only if $B$ does.

You may have noticed that definitions 4 and 5 are similarly named. This is because

Thm. $S$ is sub-finitely enumerable (definition 4) if and only if $S$ is subfinitely enumerable (definition 5).

Proof: Suppose that $S$ is sub-finitely enumerable (definition 4). Without loss of generality, suppose $S \subseteq K$ where $K$ is finitely enumerable. Take a surjection $f : [n] \to K$. Consider $S' = f^{-1}(S) \subseteq [n]$, which is clearly subfinite. Then $f|_{S'} : S' \to S$ is surjective. Therefore, $S$ is subfinitely enumerable (definition 5).

Now suppose that $S$ is subfinitely enumerable (definition 5). Without loss of generality, let $K \subseteq [n]$ and $f : K \to S$ be a surjection. Define $\sim \subseteq [n]^2$ by $a \sim b$ if and only if either $a = b$ or $a, b \in K$ and $f(a) = f(b)$. Consider the quotient set and map $\pi : [n] \to [n] / \sim$. Note that $\pi$ is surjective, demonstrating that $[n] / \sim$ is finitely enumerable.

Define $g : S \to [n] / \sim$ by $g(f(x)) = \pi(x)$. Note that $g$ is well-defined, since if $f(x) = f(y)$ then $x \sim y$ and thus $\pi(x) = \pi(y)$, so $g(f(x)) = \pi(x) = \pi(y) = g(f(y))$. And note that since $f$ is surjective, it's defined on all of $S$.

I claim that $g$ is an injection. For suppose that $g(a) = g(b)$. Write $a = f(x)$, $b = f(y)$. Then $\pi(x) = \pi(y)$. Then $x \sim y$. Then $f(x) = f(y)$. Then $a = b$.

Therefore, $S$ is sub-finitely enumerable (definition 4). $\square$

From here on out, we use sub-finitely enumerable and subfinitely enumerable interchangably.

We see that subfinitely enumerable is the definition you're looking for.

If $S$ is subfinitely enumerable, then we have an injection $f : S \to K$ where $K$ is finitely enumerable. If we have $S' \subseteq S$, then we have $f|_{S'} : S' \to K$ is also an injection, as the composition of the inclusion injection $S' \subseteq S$ and the injection $f : S \to K$. So $S'$ is subfinitely enumerable. So subfinite enumerability is closed under the taking of subsets.

If $S$ is subfinitely enumerable, then we have a surjection $f : K \to S$ where $K$ is subfinite. Suppose we have some equivalence relation $\sim$ on $S$. Then $\pi : S \to S / \sim$ is a surjection, so $\pi \circ f : K \to S / \sim$ is a surjection as the composition of surjections. Therefore, $S / \sim$ is subfinitely enumerable.

Note that "subfinitely enumerable" is therefore the smallest class $\mathbf{C}$ of sets such that (1) if $A \approx B$ and $A \in \mathbf{C}$, then $B \in \mathbf{C}$ (2) if $S \in \mathbf{C}$ and $S' \subseteq S$ then $S' \in \mathbf{C}$, (3) if $S \in \mathbf{C}$ and $\sim$ is an equivalence relation on $S$, then $S / \sim \in \mathbf{C}$, and (4) for all $n \in \mathbb{N}$, $n \in \mathbf{C}$.

Assuming the existence of power sets and of $\mathbb{N}$, the full subcategory of subfinitely enumerable sets and functions between them is essentially small. This is because every subfinitely enumerable set can be placed into a bijection with some subset of $[n] / \sim$ for some equivalence relation $\sim$ on $[n]$. So a subfinitely enumerable set is exactly a set which can be placed into bijection with some element of $\{S \mid S \subseteq [n] / \sim, n \in \mathbb{N}, \sim$ is an equivalence relation on $[n]\}$.

One final note: if you're familiar with topos theory, the proof that definitions (4) and (5) coincide can be made quite simple. For (4) implies (5), we can simply take the pullback of $[n] \to K \leftarrow S$ and use the fact that the pullback of a mono is a mono, and the pullback of an epi is an epi (in a topos). For (5) implies (4), we can take the pushout of $[n] \leftarrow k \to S$ and use the fact that the pushout of a mono is mono (in a topos), and the pushout of an epi is an epi.

  • $\begingroup$ Oh, of course, I asked the question badly! In fact, I would like to preserve both the finiteness of subsets and the finiteness of factor sets. Can I edit the question? $\endgroup$ Oct 23 at 20:16
  • $\begingroup$ @EschatumVerus Sure, but I don't know exactly what you mean by "factor sets" so you should include that in your edit. $\endgroup$ Oct 23 at 20:18
  • $\begingroup$ @MarhSaving I have edited. I meant a quotient set, sorry, for the bad translation. $\endgroup$ Oct 23 at 20:28
  • $\begingroup$ @EschatumVerus See the edit. $\endgroup$ Oct 23 at 21:29
  • $\begingroup$ @MarhSaving Thank you! I know the theory of topos, yeah. $\endgroup$ Oct 23 at 23:12

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