Proposition 2

— Let there be given an inverse system $(T_i, p_{ij})$ of topological spaces indexed by I, a topological space T, and a coherent and separating family of continuous mappings $p_i : T \to T_i$. Then:

a) For every compact subset K of T, one has $K = \bigcap_{i \in I} p_i^{-1}(p_i(K))$.

b) Let K and L be two disjoint compact subsets of T. There exists an $i \in I$ such that $p_j(K)$ and $p_j(L)$ are disjoint for $j \geq i$.

a) Let x be a point of $\bigcap_{i \in I} p_i^{-1}(p_i(K))$; for every $i \in I$, the set $K_i$ of points y in K such that $p_i(y) = p_i(x)$ is a nonempty closed subset of K. For $i \leq j$ we have $K_i \supset K_j$, and, since K is compact, the set $\bigcap_{i \in I} K_i$ is therefore nonempty. Let y be a point of $\bigcap_{i \in I} K_i$; we have $y \in K$ and $p_i(y) = p_i(x)$ for all $i \in I$, whence $y = x$; finally, $x \in K$, which proves the inclusion $K \supset \bigcap_{i \in I} p_i^{-1}(p_i(K))$; the reverse inclusion is obvious.

b) For every $i \in I$, set $M_i = p_i^{-1}(p_i(K)) \cap L$; this is a closed subset of the compact space L, we have $M_i \supset M_j$ for $i \leq j$, and, by a),

Consequently, there exists an index i such that $M_i = \varnothing$. For $j \geq i$, we have $p_j^{-1}(p_j(K)) \cap L = M_j \subset M_i = \varnothing$, whence $p_j(K) \cap p_j(L) = \varnothing$.