Set Intersection is Idempotent/Indexed Family

Theorem
Let $\langle{ F_i }\rangle_{i \mathop \in I}$ be a non-empty indexed family of sets.

Suppose that all the sets in the family are the same.

That is, suppose that for some set $S$:


 * $\forall i \in I: F_i = S$

Then $\displaystyle \bigcap_{i \mathop \in I} F_i = S$.

Proof
First we show that $\displaystyle \bigcap_{i \mathop \in I} F_i \subseteq S$.

Let $x \in \displaystyle \bigcap_{i \mathop \in I} F_i$.

Since $I$ is non-empty, it has an element $k$.

By the definition of intersection, $x \in F_k$.

By the premise, $F_k = S$, so $x \in S$.

Since this holds for all $x \in \displaystyle \bigcap_{i \mathop \in I} F_i$:


 * $\displaystyle \bigcap_{i \mathop \in I} F_i \subseteq S$

Next we show that $\displaystyle S \subseteq \bigcap_{i \mathop \in I} F_i$.

Let $x \in S$.

Then for all $i \in I$, $F_i = S$, so $x \in F_i$.

Thus by the definition of intersection: $x \in \displaystyle \bigcap_{i \mathop \in I} F_i$.

Since this holds for all $x \in S$:


 * $S \subseteq \displaystyle \bigcap_{i \mathop \in I} F_i$

By Equality of Sets: $\displaystyle \bigcap_{i \mathop \in I} F_i = S$