Let $\{S\_i\}\_{i \in I}$ be a directed projective system and $S = \varprojlim S\_i$. 


1.How to prove that $S$ is non-empty in the following case: all $S\_i$ are nonempty compact Hausdorff spaces.


2.I know that when $S\_i$ are non-empty sets and $f\_{ij}$ are surjective, $S$ may still be empty. But if we furthermore assume that $I$ is countable, then $S$ can be shown to be non-empty. What more can we say about the non-emptyness of $S$?



When is the projective limit non-empty?
This answer is to supplement Fred Rohrer's answer, copying out some details extracted from Bourbaki (in writing for instance TG.I.9.6, he was referring to a French edition of Topologie Générale, but I'll just refer to the English language edition of Bourbaki's Set Theory Treatise, starting [here](https://archive.org/stream/springer_10.1007-978-3-642-59309-3/10.1007-978-3-642-59309-3#page/n203/mode/2up), page 205/418). 


Suppose given a directed set $D$ (i.e., a partially ordered set such that any two elements have an upper bound), regarded as a category, and suppose given a functor $F: D^{op} \to Set$ whose value at $\alpha \in D$ is denoted $F\_\alpha$, with transition maps denoted $f\_{\alpha \beta}: F\_\beta \to F\_\alpha$ when $\alpha < \beta$. Suppose further that for each $\alpha \in D$, one is given a collection of subsets $S\_\alpha \subseteq P(F\_\alpha)$, each closed under arbitrary intersections [so vacuously $F\_\alpha \in S\_\alpha$!] and satisfying the finite intersection property. (For example, if the $F\_\alpha$ carry compact Hausdorff space structures, we could take $S\_\alpha$ to be the collection of closed subsets of $F\_\alpha$.) The finite intersection property could be replaced by: 



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> If $K \subseteq S\_\alpha$ is codirected, i.e., has the property that for all $A, B \in K$ there is $C \in K$ with $C \subseteq A$ and $C \subseteq B$, and if also $\emptyset \notin K$, then $\bigcap K \neq \emptyset$. (\*\*) 
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**Theorem:** Assume that $f\_{\alpha\beta}(C) \in S\_\alpha$ whenever $C \in S\_\beta$, and that $f^{-1}(x) \in S\_\beta$ whenever $x \in F\_\alpha$. Then for the projective limit $L$ with universal cone $f\_\alpha: L \to F\_\alpha$, we have $f\_\alpha(L) = \bigcap\_{\alpha \leq \beta} f\_{\alpha\beta}(F\_\beta)$. 


Note we have the following corollary. 


**Corollary:** Under the assumptions of the theorem, if all the $F\_\alpha$ are nonempty, then so is $L$. 


This follows from the theorem by choosing any $\alpha \in D$ and taking $K = \{f\_{\alpha\beta}(F\_\beta): \alpha \leq \beta\}$, noting that $K$ is codirected because $D$ is directed, and applying the highlighted condition (\*\*). (In the exceptional case where $D$ is empty, the projective limit $L$ is the terminal set, hence nonempty.) 


Here is an outline of how Bourbaki proves the theorem. First he sets up a Zorn's lemma situation: let $\Sigma$ be the poset whose elements are tuples of the form $(A\_\alpha)\_{\alpha \in D}$ such that $A\_\alpha \in S\_\alpha$ and $A\_\alpha \neq \emptyset$ and $f\_{\alpha\beta}(A\_\beta) \subseteq A\_\alpha$ whenever $\alpha \leq \beta$, and with $\Sigma$ partially ordered by $(A\_\alpha) \preceq (B\_\alpha)$ if $B\_\alpha \subseteq A\_\alpha$ for all $\alpha \in D$. (Note the change of direction.) Using the highlighted form of the finite intersection property, he notes that every chain in $(\Sigma, \preceq)$ has an upper bound. Thus $\Sigma$ has a maximal element, by Zorn's lemma. 


He next shows that if $(A\_\alpha)$ is maximal, then $f\_{\alpha\beta}(A\_\beta) = A\_\alpha$. In effect he considers $A\_{\alpha}' := \bigcap\_{\alpha \leq \beta} f\_{\alpha\beta}(A\_\beta)$ and checks that $(A\_{\alpha}')$ satisfies the requirements for belonging to $\Sigma$ and that $f\_{\alpha\beta}(A\_{\beta}') = A\_{\alpha}'$. Then observe $A\_{\alpha}' \subseteq A\_\alpha$, so $A\_{\alpha}' = A\_\alpha$ by maximality of $(A\_\alpha)$. 


Next he shows that each $A\_\alpha$ is a singleton. Hold fixed an $\alpha$, select $x\_\alpha \in A\_\alpha$, and put $B\_\beta := A\_\beta \cap f^{-1}\_{\alpha\beta}(x\_\alpha)$ if $\alpha \leq \beta$; otherwise put $B\_\beta = A\_\beta$. Bourbaki shows that $(B\_\beta)\_{\beta \in D}$ belongs to $\Sigma$, and clearly $B\_\beta \subseteq A\_\beta$ for all $\beta \in D$, so $A\_\beta = B\_\beta$ for all $\beta$ by maximality of $(A\_\beta)$, and in particular $A\_\alpha = B\_\alpha = \{x\_\alpha\}$. 


Finally, he shows $f\_\alpha(L) = \bigcap\_{\alpha \leq \beta} f\_{\alpha\beta}(F\_\beta)$. The direction $\subseteq$ is immediate. For the other direction, suppose $x\_\alpha \in \bigcap\_{\alpha \leq \beta} f\_{\alpha\beta}(F\_\beta)$, and put $B\_\beta := f\_{\alpha\beta}^{-1}(x\_\alpha)$ if $\alpha \leq \beta$, else put $B\_\beta = F\_\beta$. One argues that $(B\_\beta)\_{\beta \in D}$ belongs to $\Sigma$ and that there is a maximal element $(A\_\beta)$ that is $\preceq$-above $(B\_\beta)$, with $A\_\beta = \{y\_\beta\}$. The tuple $y = (y\_\beta)\_{\beta \in D}$ belong to the limit $L$, and $f\_\alpha(y) = y\_\alpha = x\_\alpha$, as was to be shown. $\Box$ 


As for question 2. of the OP, there is the slightly more general statement that if the directed set $D$ admits a countable cofinal subset and all the $f\_{\alpha\beta}$ are surjective, then the limit is nonempty (this is without all the extra baggage of the $S\_\alpha$ above). The countable cofinal set allows us to reduce to the case $D = \mathbb{N}$ without loss of generality, and then from there it's an easy application of dependent choice. See Proposition 5 in Bourbaki, just before the theorem we quoted above. 


