Lemma 5.14.6. Let $\mathcal{I}$ be a cofiltered category and let $i \mapsto X_ i$ be a diagram over $\mathcal{I}$ in the category of topological spaces. If each $X_ i$ is quasi-compact, Hausdorff, and nonempty, then $\mathop{\mathrm{lim}}\nolimits X_ i$ is nonempty.
Proof. In the proof of Lemma 5.14.5 we have seen that $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is the intersection of the closed subsets
\[ Z_\varphi = \Gamma _\varphi \times \prod \nolimits _{l \not= j, k} X_ l \]
inside the quasi-compact space $\prod X_ i$ where $\varphi : j \to k$ is a morphism of $\mathcal{I}$ and $\Gamma _\varphi \subset X_ j \times X_ k$ is the graph of the corresponding morphism $X_ j \to X_ k$. Hence by Lemma 5.12.6 it suffices to show any finite intersection of these subsets is nonempty. Assume $\varphi _ t : j_ t \to k_ t$, $t = 1, \ldots , n$ is a finite collection of morphisms of $\mathcal{I}$. Since $\mathcal{I}$ is cofiltered, we can pick an object $j$ and a morphism $\psi _ t : j \to j_ t$ for each $t$. For each pair $t, t'$ such that either (a) $j_ t = j_{t'}$, or (b) $j_ t = k_{t'}$, or (c) $k_ t = k_{t'}$ we obtain two morphisms $j \to l$ with $l = j_ t$ in case (a), (b) or $l = k_ t$ in case (c). Because $\mathcal{I}$ is cofiltered and since there are finitely many pairs $(t, t')$ we may choose a map $j' \to j$ which equalizes these two morphisms for all such pairs $(t, t')$. Pick an element $x \in X_{j'}$ and for each $t$ let $x_{j_ t}$, resp. $x_{k_ t}$ be the image of $x$ under the morphism $X_{j'} \to X_ j \to X_{j_ t}$, resp. $X_{j'} \to X_ j \to X_{j_ t} \to X_{k_ t}$. For any index $l \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ which is not equal to $j_ t$ or $k_ t$ for some $t$ we pick an arbitrary element $x_ l \in X_ l$ (using the axiom of choice). Then $(x_ i)_{i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})}$ is in the intersection
\[ Z_{\varphi _1} \cap \ldots \cap Z_{\varphi _ n} \]
by construction and the proof is complete. $\square$
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