Is inverse limit topology closed in product topology?

Question

Suppose $\{X_i\}_{i\in I}$ is an inverse system of topological spaces. Denote $X=\varprojlim X_i$. It's naturally a subspace of the product topology $\prod_{i} X_i$. I want to ask: is $X$ closed in the product topology? If not could you share a counterexample? Thanks ahead.

Answer 1

In general it is not closed, but it is closed in the case of the $X_i$ being Hausdorff:

1. As an easy counterexample take $I = \{1,2\}$ all $X_i$ to be a non-Hausdorff space $X$, then $\varprojlim X_i \subset \prod X_i$ is the diagonal $\Delta X\subset X\times X$ and maybe you are familiar with the characterization of Hausdorff being that the diagonal is closed, as this is an equivalence, our inverse limit can't be closed. (See https://en.wikipedia.org/wiki/Hausdorff_space)

The proof given here is in a way a generalization of the proof of the equivalence of Hausdorff and the closure of the diagonal in the product:

2. Claim: Given a projective system $\{X_i,\phi_{ij},I\}$ of Hausdorff spaces then its limit $\varprojlim X_i$ is a closed subspace of $\prod_{i\in I} X_i$. (Projective System in this case is equivalent to inverse system)

Take $(x_i)_I \in \prod_{i\in I} X \setminus \varprojlim X_i$ then there are $j\preceq k \in I$ such that $\phi_{kj}(x_k)\not=x_j$, as $X_j$ is Hausdorff we can choose disjoint neighborhoods $U_j,V_j\subset X_j$ of $X_j, \phi_{kj}(x_k)$ respectively. Now $V_k = \phi_{kj}^{-1}(V_j) \subset X_k$ is a neighborhood of $x_k$. Therefor $\prod_{i\in I} W_i$ with $W_i = X_i$ for $i\not=k,j$ and $W_k = V_k, W_j = U_j$ is a neighborhood of $(x_i)_I$ and disjoint from $\varprojlim X_i$ as $\phi_{kj}(W_k)$ disjoint from $W_j$. So $\varprojlim X_i$ is closed as the complement of an open set.

As a reference see for example: 1. chapter of Luis Ribes and Pavel Zalesskii. Profinite groups. English. 2nd ed. Vol. 40. Ergeb. Math. Grenzgeb., 3. Folge. Berlin: Springer, 2010. isbn: 978-3-642-01641-7