# The empty set as an Indexing set. [duplicate]

For each $\alpha\in I$, let $A_\alpha$ be a subset of some nonempty set $S$. So if $I=\emptyset$, then $$\bigcup_{\alpha\in I} A_\alpha=\emptyset$$ and $$\bigcap_{\alpha\in I} A_\alpha=S.$$

Why is this true?

• Write down the definition of union and intersection. Simplify. – user14972 Dec 21 '14 at 21:33
• – Henno Brandsma Dec 21 '14 at 21:35

$x \in \bigcup_{\alpha \in I} A_\alpha$, by definition means that there exists some $\alpha \in I$ such that $x \in A_\alpha$. So if $I$ is empty this cannot be true for any $x$ (as here is no $\alpha$), so the union is empty.
The last one is more "controversial". It ensures that de Morgan also holds for complements relative to $S$, which is our "universe" in this case: $S \setminus \bigcap_{\alpha \in I} A_\alpha = \bigcup_{\alpha \in I} (S \setminus A_\alpha) = \emptyset$ by the previous. So the intersection "should" equal $S$. Or otherwise put: $x \in \bigcap_{\alpha \in I} A_\alpha$ iff for all $\alpha \in I$ it holds that $x \in A_\alpha$. If $I = \emptyset$, then there can be no counterexample (where $x \notin A_\alpha$ and $\alpha \in I$), as there are no elements in $I$ to make a counterexample with. So this holds for all $x$. And as our "universe of discourse" is $S$, we conventially say this set equals $S$ (but it could have equaled any superset of $S$ as well, if we consider all $A_\alpha$ to be a subset of that!).