Cofiltered limit of $T_0$ and quasi compact spaces is again quasi compact.

Question

Let $\mathscr{I}$ be a cofiltered category and consider a cofiltered diagram of topological spaces $X_i$ with each space being quasi compact and $T_0$. For every $X_{i} \xrightarrow{f_{i,j}} X_j$ morphism in the diagram we have that $f_{i,j}$ is a quasi compact map. That is, if $U \subset X_j$ is open and quasi compact then its preimage is also open and quasi compact.

I would like to prove that the limit of diagram, let's called it $X$ is quasi compact. For the moment I am not getting very far. I am not even able to prove that $X$ is non empty.

I don't know if it is useful but every $X_i \cong \operatorname{Spec} R_i $ for some commutative ring.

Answer 1

I found relevant counterexamples in this paper. I'll repeat them here:

Example 2 on p. 5: Take $\varprojlim_{n\in \mathbb{N}} X_n$, where each $X_n$ is $\mathbb{N}$ with the cofinite topology, and the connecting map $f_{n+1}\colon X_{n+1}\to X_n$ is $f_{n+1}(k) = k+1$. The limit is empty.

Example 3 on p. 6: Take $\varprojlim_{n\in\mathbb{N}} X_n$, where $X_n$ is $\mathbb{N}$ with the topology generated by the cofinite sets and the singletons $\{0\},\dots,\{n-1\}$, and the connecting map $f_{n+1}\colon X_{n+1}\to X_n$ is $f_{n+1}(k) = k$. The limit is $\mathbb{N}$ with the discrete topology, which is not compact.

A trick in both of these examples is that the topologies are all Noetherian, so every subset is compact, and your condition that the connecting maps are compact trivializes.

But under the additional assumption that the $X_i$ are spectra of commutative rings, I can answer your question by some abstract nonsense:

• The spectra of commutative rings are spectral spaces: they are sober (every irreducible closed set has a unique generic point), and their compact open sets form a basis for the topology which is closed under finite intersections.
• The spectral spaces are exactly the cofiltered limits (in the category of topological spaces) of finite sober spaces: see the Stacks project for a proof.
• Now it's a fact that the compact maps between spectral spaces are exactly the ones which factor through presentations of the spaces as cofiltered limits of finite sober spaces, i.e. they're the maps in the category pro-(Finite Sober Spaces). This follows from the Stone duality between the category of spectral spaces (together with compact maps) and the category of distributive lattices, since every map of distributive lattices factors through presentations as filtered colimits of finite distributive lattices, i.e. DistLat = ind-FinDistLat.
• Summing up, we can replace our cofiltered limit of spectral spaces with a big cofiltered limit of finite sober spaces, proving that a cofiltered limit of spectral spaces is a spectral space, and in particular it's compact.

No doubt there's a more direct way to see it, but this is the abstract approach that occurred to me.