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It is well known that $\mathbf{Set}$ is an $\aleph_0$-accessible category, but I'm very inexperienced and I'm not sure how to prove it in detail. In particular, I need to find a set $\Omega$ of finitely presentable objects (sets) such that every other set is a directed colimit of elements of $\Omega$.

Here is my attempt. Given any set $A$, we can define a directed diagram $([A]^{\aleph_0},\subseteq)$, where $[A]^{\aleph_0}$ denotes the set of finite subsets of $A$. If we define $D:([A]^{\aleph_0},\subseteq)\to\mathbf{Set}$ by $D(S)=S$ and $D(S_1\to S_2)=i_{12}$, where $i_{12}$ is the inclusion from $S_1$ to $S_2$, then $(S\stackrel{i_S}{\longrightarrow} A)_{S\in [A]^{\aleph_0}}$ is a colimit of $D$, where $i_S$ is the inclusion from $S$ to $A$.

In addition, we consider any diagram $F:(I,\leq)\to\mathcal{K}$ with colimit $(F_i\stackrel{\eta_i}{\longrightarrow} \text{colim}( F))_{i\in I}$ and we take some object $C$ isomorphic to $F_j$ (with isomorphism $h:C\to F_j$), for some $j\in I$. Then, the diagram $F':(I,\leq)\to\mathcal{K}$ given by $F'(j)=C$, $F'(i)=F(i)$ if $i\not=j$ and $F'(i\to j)=h^{-1}\circ F(i\to j)$, $F'(j\to i)=F(j\to i)\circ f$ has a colimit $(F_i'\stackrel{\mu_i}{\longrightarrow} \text{colim}( F'))_{i\in I}$ given by $\text{colim}( F')=\text{colim}( F)$ and $\mu_i=\eta_i$ if $i\not=j$ and $\mu_j=\eta_j\circ h$. Basically, we are replacing an object in the diagram and $\text{colim}(F)$ does not change. Is this correct?

Finally, we can replace $D(S)$ with $n$, where $n=|S|$, for every $S\in [A]^{\aleph_0}$. Since $n$ is isomorphic in $\mathbf{Set}$ to any set with cardinality $n$, the previous paragraph implies that the resulting diagram $D'$ (after replacing every object with its cardinality), has $\text{colim}(D')=\text{colim}(D)=A$. Then, this proves that any set $A$ is the directed colimit of sets in $\omega$ (the set of natural numbers), which is waht we wanted to prove.

Is the above reasoning correct? If not, where did I go wrong? And if yes, isn't it a little strange that an uncountable set can be constructed from a diagram with only the natural numbers, which is a countable set?

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Yes, you can replace the values of the functor $D$ by the finite cardinals. (Remember that in a colimit diagram, you're allowed to use the same object multiple times!) Thus what you did is correct.

To make the fact that you constructed a way to write an uncountable set as a colimit of finite cardinals more digestable, remember that a directed colimit in $\mathbf{Set}$ is just a directed union, so what we are doing is writing $A=\bigcup_{X \in [A]^{\aleph_0}} X$, i.e. we're just saying that every set is the union of its finite subsets. Now because we can use the same object more than once in the diagram we're taking the colimit over, we can replace this diagram with one that has only objects in $\omega$. If $A$ is uncountable, then the diagram will contain uncountably many copies of $1$, for example.

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  • $\begingroup$ My problem and confusion does not lie in those points. The point is that the definition of $\lambda$-accessible category, as I understand it, requires a fixed set $\Omega$ of $\lambda$-presentable objects such that every object is a $\lambda$-directed colimit of elements of that set. The colimit of $D$ is $A$, but it is clear that $D(i)$ may not be in $\Omega$. If the set $\Omega$ depends on the object, then my doubt would be clarified. $\endgroup$
    – Petersu
    Feb 25 at 18:24
  • $\begingroup$ @Petersu sorry for misunderstanding you. Yes, you can replace the values of $D$ by elements in $\omega$, as long as you don't change the index category (which is what I thought you were doing), then this doesn't change the colimit. $\endgroup$ Feb 25 at 18:26
  • $\begingroup$ It is true that I did not express my doubt vey well. But then, if my approach is correct, how can I digest the fact that I can construct an uncountable set with a diagram that only has elements in $\omega$? $\endgroup$
    – Petersu
    Feb 25 at 18:31
  • $\begingroup$ @Petersu I have edited. $\endgroup$ Feb 25 at 18:36
  • $\begingroup$ Thank you very much, it is very helpful for me that you assure me that what I have done is right, because I am learning on my own. It still seems very strange to me, but that is a problem of the human mind. $\endgroup$
    – Petersu
    Feb 25 at 18:45

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