Proposition 3

Let $(A_i)_{i \in I}$ be a family of subsets of a topological space $X$, such that one of the following properties holds: a) The interiors of the $A_i$ cover $X$. b) $(A_i)_{i \in I}$ is a locally finite closed covering of $X$ (§ 1, no. 5).

Under these conditions, a subset $B$ of $X$ is open (resp. closed) in $X$ if and only if each of the sets $B \cap A_i$ is open (resp. closed) in $A_i$.

Clearly if $B$ is open (resp. closed) in $X$, then $B \cap A_i$ is open (resp. closed) in $A_i$. Conversely, suppose first that condition a) is satisfied; since $(\complement B) \cap A_i = A_i - (B \cap A_i)$, it is enough, by duality, to consider the case in which each of the $B \cap A_i$ is open with respect to $A_i$. In this case $B \cap \dot{A}_i$ is open in $\dot{A}_i$ for each $i \in I$, and therefore open in $X$; and since $B = \bigcup_i (B \cap \dot{A}_i)$ by hypothesis, it follows that $B$ is open in $X$.

Now suppose that b) is satisfied; by duality again, we need only consider the case in which each of the $B \cap A_i$ is closed in $A_i$, and therefore closed in $X$. Since the family $(B \cap A_i)$ is locally finite and $B = \bigcup_i (B \cap A_i)$, it follows from § 1, no. 5, Proposition 4 that $B$ is closed in $X$.