I am doing exercise 1.9 from Lenstra's Galois theory for Schemes.
Let $\left((S_i)_{i\in I},I,f_{ij}\right)$ be a projective system, where $I$ is countable, all $S_i$ are non-empty and all $f_{ij}$ are surjective.
Prove that $\varprojlim_{i\in I}S_i\neq \varnothing.$
(I will assume he means countably infinite, the finite case being trivial.) Once we can assume without loss of generalisation that $I=\mathbf{N}$, this becomes easy. The crux of the problem, if we don't want to make our hands too dirty, lies in the justification, I think.
Apparently, we can not make this assumption immediately, since a directed countable set is not necessarily totally ordered. Can someone give an example of this?
I know the fact that $\varprojlim_{i\in I}S_i\simeq\varprojlim_{j\in J}S_j$, where $J\subset I$ is a cofinal subset. I saw this question. This seems to prove the following: if $I$ is countable, then there exists a cofinal totally ordered subset. He/she proves it in the following way:
If $I$ is countable, there is a bijection $I\simeq \mathbf{N}$, so we can write $I=\{a_n \mid n\in \mathbf{N} \}$. Now define a subset $J=\{x_n \mid n\in\mathbf{N} \}$ of $I$ as follows: $x_0=a_0$, $x_{n+1}=a_{i_n}$, where $i_n:=\min \{j\in \mathbf{N}\mid a_j\geqslant x_n,a_j\geqslant a_n \}$ (this minimum exists because $I$ is directed). This gives $x_{n+1}\geqslant x_n$ and (2) $x_{n+1}\geqslant a_n$, which shows respectively that $J$ is totally ordered and that $J\subset I$ is cofinal.
How does this help? Is there some theorem I don't know of that says that a totally ordered directed countable set is necessarily $\mathbf{N}$? My set theory is pretty rusty apparently. Any help is very much appreciated. :)
Of course that there are countable directed sets which are not $\Bbb N$. For example $\Bbb{Z,Q}$ and even $\Bbb N$ ordered by $|$ which is not totally ordered, but it is in fact a lattice. The same can be said on the finite subsets of $\Bbb N$ ordered by inclusion.
But nevertheless, if $I$ is a countable directed set, it contains a subset which is totally ordered and it is cofinal. This is not necessarily for in general, for example the finite subsets of $\Bbb R$ ordered by inclusion. Any chain must be countable, and its limit must also be countable.