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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?

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$x \in \bigcup_{\alpha \in I} A_\alpha$, 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!).

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