# Is every set a filtered colimit of finite sets?

Is the following statement correct in the category of sets?

Let $$X$$ be any set. Then there exists a filtered small category $$I$$ and a functor $$F:I\to \mathrm{Set}$$ such that for all $$i\in I$$ the set $$F(i)$$ is finite, and such that $$X \; = \; \mathrm{colim}_{i\in I} F(i) .$$

Are there references on results of this type in the literature?

## 2 Answers

The answer is yes: every set is the union of its finite subsets.

So take $$I = P_{\text{finite}}(X)$$ with as morphisms the inclusion maps, and $$F : I \to \text{Set}$$ the inclusion.

One answer already mentions the diagram of finite subsets of $$X$$. You would have to check that taking the union of this system actually is the colimit (which is an easy exercise).

Since you asked for a reference, Locally Presentable and Accessible Categories by J. Adámek and J. Rosický is a great book on this kind of stuff. In particular example 1.2(1) already mentions the diagram of finite subsets.