ThmDex – An index of mathematical definitions, results, and conjectures.
Result R2079 on D76: Set intersection
Isotonicity of set intersection

Let $E_j$ and $F_j$ each be a D11: Set for every $j \in J$ such that
 (i) $$\forall \, j \in J : E_j \subseteq F_j$$
Then $$\bigcap_{j \in J} E_j \subseteq \bigcap_{j \in J} F_j$$
Proofs
Proof 0
Let $E_j$ and $F_j$ each be a D11: Set for every $j \in J$ such that
 (i) $$\forall \, j \in J : E_j \subseteq F_j$$
If the intersection $\bigcap_{j \in J} E_j$ is empty, the claim follows from R7: Empty set is subset of every set. Suppose thus that $\bigcap_{j \in J} E_j$ is not empty and fix $x \in \bigcap_{j \in J} E_j$. By definition of set intersection, now $x$ belongs to $E_j$ for every $j \in J$. Combining this with the initial assumptions, one then has $$x \in E_j \subseteq F_j$$ and thus $x \in F_j$ for every $j \in J$. Therefore, again by the definition of a set intersection, $x \in \bigcap_{j \in J} F_j$. Since $x \in \bigcap_{j \in J} E_j$ was arbitrary, we have the inclusion $\bigcap_{j \in J} E_j \subseteq \bigcap_{j \in J} F_j$. $\square$