It was probably asked multiple times before which is about basic logic. But I couldnt find any.
let X a set and I is index set. for $i\in I$, $A_i$ is a subset of X. if $I=\emptyset$, how can we show
$$\bigcup_{i\in \emptyset} A_i=\emptyset\quad\text{and}\quad\bigcap_{i\in \emptyset}A_i=X$$
is it because from the definition if $x\in\bigcap_{i\in \emptyset}A_i\iff\forall i(x\in \emptyset\Rightarrow x\in A_i) $ which equals to $\forall i(x\notin \emptyset \lor x\in A_i)$ which is true for all $x\in X$ because $x\notin \emptyset$ is always true.
similarly if $x\in\bigcup_{i\in \emptyset} A_i\iff\exists i(i\in \emptyset \land x\in A_i) $ right statement is wrong because $i\in \emptyset$ is wrong always. that is, for all $x \in X, x\notin\bigcup_{i\in \emptyset} A_i$ so $\bigcup_{i\in \emptyset} A_i=\emptyset$
is this sufficient?
