A finite object in Set is a finite Set I'd like to prove that an object of $\bf{Set}$ is finite iff it is a finite set. Finite object is defined as being small with respect to a finite cardinal; here's the definition of smallness I'm assuming:
"Let $k$ be a cardinal and $\mathscr{C}$ a cocomplete category. An object $A \in \mathscr{C}$ is said to be $\textit{$k$-small}$ iff for all $k$-filtered ordinal $\lambda$ (which means that $cof(\lambda) >k$ strictly) and for all  colimit preserving-functors $X: \lambda \to \mathscr{C}$ we have that the canonical morphism of sets $$colim_{\beta<\lambda}\mathscr{C}(A,X_{\beta}) \to \mathscr{C}(A,colim_{\beta < \lambda} X_{\beta})$$ is an isomorphism".
I'm able to prove that "finite set$\implies$finite object", but the converse gives me some problems. I tried to suppose that it's an infinite set and to use it's cardinality as an infinite cardinal that gives an absurd with such an iso, but I got stuck..
Any help or any hint is welcome! Thanks in advance :)
 A: I'm missing something or you only lack the easy side of [LPAC] Example 1.2.1?
Every set is the directed colimit of its finite subsets. Hence if $K$ is a finite object, then you must have that $1_K\colon K\to K=\varinjlim{}_{F\in \mathcal P_0(K)} F$ must factor through one of the inclusions of a finite subset of $K$; hence $K$ is a finite set.
A: Could be the following a naïve counter-example? That is, an infinite set which is not small.
Even better: could it be generalized as to prove "infinite set" $\Longrightarrow$ "non finite object"?
I don't know: I'm just asking. So, consider this just as a question, more than an answer if you whish.
Take as $\mathcal{C}$ the category of sets, $\mathbf{Set}$, $A = \mathbb{N}$ and $X : \mathbb{N} \longrightarrow \mathbf{Set}$ the functor $X_n = \{ 1, \dots , n \}$. Then, on the left handside of your canonical morphism you would have:
$$
\mathrm{colim}_n \mathbf{Set}\ (\mathbb{N}, X_n) = \mathrm{colim}_n \{1, \dots , n \}^{\mathbb{N}} = \bigcup_n \{1, \dots , n \}^{\mathbb{N}}  \ .
$$
Here, $\{1, \dots , n \}^{\mathbb{N}} $ is the set of all sequences you can make out of the first $n$ natural numbers.
On the other hand, at the other end of your canonical morphism, you would have:
$$
\mathbf{Set}(\mathbb{N}, \mathrm{colim}_n X_n) = \mathbf{Set}(\mathbb{N}, \mathbb{N}) = \mathbb{N}^{\mathbb{N}}  \ .
$$
That is, the set of all sequences of natural numbers.
Then, your canonical map cannot be a bijection. Namely, the sequence of natural numbers $(n) = (1, 2, 3, \dots , n, \dots )$ belongs to $\mathbb{N}^{\mathbb{N}}$, obviously, but doesn't belong to the set on the left hand side, since it doesn't belong to any of the subsets $ \{1, \dots , n \}^{\mathbb{N}} $.
