Corollary

Let $A$ be a subset of $X$ and let $A_\alpha$ denote $f_\alpha(A)$ for each $\alpha \in I$. Then:

(i) The $A_\alpha$ (resp. the $\overline{A}_\alpha$) form an inverse system of subsets of the $X_\alpha$, and

(ii) If $A$ is closed in $X$, $A = \varprojlim A_\alpha = \varprojlim \overline{A}_\alpha$.

The first assertion of (i) follows from the relations $f_\alpha = f_{\alpha\beta} \circ f_\beta$ for $\alpha \leq \beta$ and from the continuity of the $f_{\alpha\beta}$ (§ 2, no. 1, Theorem 1). Let $A'$ denote

then it is clear that $A'$ is closed and contains $A$, so that $\overline{A} \subset A'$. Conversely, let $x \in A'$; we have to show that $x$ lies in the closure of $A$. By virtue of Proposition 9 it is enough to prove that every neighbourhood of $x$ which is of the form $\overline{f}_\alpha(U_\alpha)$, with $\alpha \in I$ and $U_\alpha$ open in $X_\alpha$, meets $A$. Now, by hypothesis, $f_\alpha(x) \in U_\alpha$, and since $f_\alpha(x) \in \overline{A}$ we have $U_\alpha \cap A_\alpha \neq \varnothing$, which means that $A \cap \overline{f}_\alpha(U_\alpha)$ is not empty.

To establish (ii) it is enough to remark that, without any restriction on $A$, we have $A \subset \lim A_\alpha \subset \lim \overline{A}_\alpha$; now if $A$ is closed, then from (i)

and (ii) follows.