Question

Then one can construct a model for the inverse limit by taking all the compatible sequences. This is a subspace of a product of compact spaces. This product is compact by Tychonoff. If all the spaces are Hausdorff, then this is even a closed subspace.

However, if the spaces are not Hausdorff, it needn't be a closed subspace. If you take a two point space with the trivial topology as $X_n$ and constant structure maps, you will get as the inverse limit the space of all constant sequences, which is not a closed subspace of the infinite product, as the infinite product also has the trivial topology.

But the space is again compact. So I am wondering, whether there is a generalization of the proof of Tychonoff's theorem, that applies directly to inverse limits.

Answer 1

What does this example do ...

All spaces are on set $\{1,2,\dots\}$. Space $X_n$ has topology that makes $\{1,2,\dots,n\}$ discrete and $\{n+1,\dots\}$ indiscrete. Of course $X_n$ is compact non-Hausdorff. Map $X_{n+1} \to X_n$ by the "identity". Inverse limit is ... ???

Answer 2

Although the question is quite old I just ran into it. You might still want to (if you haven´t done so yet) look at the article:

A.H. Stone, Inverse limits of compact spaces, General Topology and its Applications (1979).

It contains some positive results.

Answer 3

Let $X'=\{\frac1n:n\in\mathbb N\}\subset\mathbb R$, with its subspace topology, and let $X=X\cup\{0_1,0_2\}$ be the result of adding two distinct limits to the obvious sequence; $X$ is a compact non-Hausdorff space. Let $s:X\to X$ be the map which swaps the two zeroes, which is continuous. Then the equalizer of $s:X\to X$ and $\mathrm{id}_X:X\to X$ is $X'$, which is not compact. If in 'inverse limit' you include general limits, then the answer is thus no.