Lemma 5.24.1. Let $\mathcal{I}$ be a category. Let $i \mapsto X_ i$ be a diagram of spectral spaces such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral.

1. Given subsets $Z_ i \subset X_ i$ closed in the constructible topology with $f_ a(Z_ j) \subset Z_ i$ for all $a : j \to i$ in $\mathcal{I}$, then $\mathop{\mathrm{lim}}\nolimits Z_ i$ is quasi-compact.

2. The space $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is quasi-compact.

Proof. The limit $Z = \mathop{\mathrm{lim}}\nolimits Z_ i$ exists by Lemma 5.14.1. Denote $X'_ i$ the space $X_ i$ endowed with the constructible topology and $Z'_ i$ the corresponding subspace of $X'_ i$. Let $a : j \to i$ in $\mathcal{I}$ be a morphism. As $f_ a$ is spectral it defines a continuous map $f_ a : X'_ j \to X'_ i$. Thus $f_ a|_{Z_ j} : Z'_ j \to Z'_ i$ is a continuous map of quasi-compact Hausdorff spaces (by Lemmas 5.23.2 and 5.12.3). Thus $Z' = \mathop{\mathrm{lim}}\nolimits Z_ i$ is quasi-compact by Lemma 5.14.5. The maps $Z'_ i \to Z_ i$ are continuous, hence $Z' \to Z$ is continuous and a bijection on underlying sets. Hence $Z$ is quasi-compact as the image of the surjective continuous map $Z' \to Z$ (Lemma 5.12.7). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).