Interpretation of closure in inverse limit

Question

Can one interpret the closure of a set inside an inverse limit as the closure of its individual components? I have not been able to find a source confirming or denying this claim. I have only been able to show that there exist closed sets in the inverse limit that have an open component, making me believe that it is not true (since then we may have equality $S= \overline{S}$ where $S$ has an open component and $\overline{S}$ does not. Though perhaps these are identified in the limit).

To be more explicit, suppose $X = \varprojlim X_i$ is an inverse limit of topological spaces, and $S= \varprojlim S_i \subseteq X$. Does it hold that $\overline{S} = \varprojlim \overline{S_i}$?

Thanks in advance as ever,

M

Answer 1

It is not completely clear what you want to know.

For notation let $f^j_i: X_j \to X_i$ be the bonding maps of the inverse system and $\pi_i : \varprojlim X_i \to X_i$ the projections.

Variant 1.

Let $S \subset \varprojlim X_i$ and $S_i = \pi_i(S)$. Then $f^j_i(S_j) \subset S_i$ and by continuity we get $f^j_i(\overline S_j) \subset \overline{f^j_i( S_j)} \subset \overline S_i$. One can show that $$\overline S = \varprojlim \overline S_i .$$

As commented by Stephan, a proof can be found in

Bourbaki, General Topology, Chapter 1, §4, Corollary to Proposition 9

Also see Proposition 2.5.6 in

Engelking, R. (1989). General topology. Sigma series in pure mathematics, 6.

Variant 2.

Let $S_i \subset X_i$ be subpaces such that $f^j_i(S_j) \subset S_i$. By continuity we get $f^j_i(\overline S_j) \subset \overline{f^j_i( S_j)} \subset \overline S_i$.

We have $$\varprojlim \overline{S_i} = \{(x_i )\in \prod \overline{S_i} \mid f^j_i(x_j) = x_i \text{ for all } j \ge i \} = \varprojlim X_i \cap \prod \overline{S_i} .$$ Since $\prod \overline{S_i}$ is closed in $\prod X_i$, we see that $\varprojlim \overline{S_i}$ is closed in $\varprojlim X_i$. Hence $$\overline{\varprojlim S_i} \subset \varprojlim \overline{S_i} .$$

However, in general it is not true that $$\overline{\varprojlim S_i} = \varprojlim \overline{S_i} .$$ See Inverse Limit of Dense Subsets is Dense for examples.