Lemma 4.21.7. If $S : \mathcal{I} \to \textit{Sets}$ is a cofiltered diagram of sets and all the $S_ i$ are finite nonempty, then $\mathop{\mathrm{lim}}\nolimits _ i S_ i$ is nonempty. In other words, the limit of a directed inverse system of finite nonempty sets is nonempty.

Proof. The two statements are equivalent by Lemma 4.21.5. Let $I$ be a directed set and let $(S_ i)_{i \in I}$ be an inverse system of finite nonempty sets over $I$. Let us say that a subsystem $T$ is a family $T = (T_ i)_{i \in I}$ of nonempty subsets $T_ i \subset S_ i$ such that $T_{i'}$ is mapped into $T_ i$ by the transition map $S_{i'} \to S_ i$ for all $i' \geq i$. Denote $\mathcal{T}$ the set of subsystems. We order $\mathcal{T}$ by inclusion. Suppose $T_\alpha$, $\alpha \in A$ is a totally ordered family of elements of $\mathcal{T}$. Say $T_\alpha = (T_{\alpha , i})_{i \in I}$. Then we can find a lower bound $T = (T_ i)_{i \in I}$ by setting $T_ i = \bigcap _{\alpha \in A} T_{\alpha , i}$ which is manifestly a finite nonempty subset of $S_ i$ as all the $T_{\alpha , i}$ are nonempty and as the $T_\alpha$ form a totally ordered family. Thus we may apply Zorn's lemma to see that $\mathcal{T}$ has minimal elements.

Let's analyze what a minimal element $T \in \mathcal{T}$ looks like. First observe that the maps $T_{i'} \to T_ i$ are all surjective. Namely, as $I$ is a directed set and $T_ i$ is finite, the intersection $T'_ i = \bigcap _{i' \geq i} \mathop{\mathrm{Im}}(T_{i'} \to T_ i)$ is nonempty. Thus $T' = (T'_ i)$ is a subsystem contained in $T$ and by minimality $T' = T$. Finally, we claim that $T_ i$ is a singleton for each $i$. Namely, if $x \in T_ i$, then we can define $T'_{i'} = (T_{i'} \to T_ i)^{-1}(\{ x\} )$ for $i' \geq i$ and $T'_ j = T_ j$ if $j \not\geq i$. This is another subsystem as we've seen above that the transition maps of the subsystem $T$ are surjective. By minimality we see that $T = T'$ which indeed implies that $T_ i$ is a singleton. This holds for every $i \in I$, hence we see that $T_ i = \{ x_ i\}$ for some $x_ i \in S_ i$ with $x_{i'} \mapsto x_ i$ under the map $S_{i'} \to S_ i$ for every $i' \geq i$. In other words, $(x_ i) \in \mathop{\mathrm{lim}}\nolimits S_ i$ and the lemma is proved. $\square$

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