Set Intersection Preserves Subsets

Theorem

Let $A, B, S, T$ be sets.

Then:

$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

Corollary

Let $A, B, S$ be sets.

Then:

$A \subseteq B \implies A \cap S \subseteq B \cap S$

Families of Sets

Let $I$ be an indexing set.

Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.

Let:

$\forall \beta \in I: A_\beta \subseteq B_\beta$

Then:

$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$

Proof

Let $A \subseteq B$ and $S \subseteq T$.

Then:

 $\ds x \in A$ $\implies$ $\ds x \in B$ Definition of Subset $\ds x \in S$ $\implies$ $\ds x \in T$ Definition of Subset

Now we invoke the Praeclarum Theorema of propositional logic:

$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$

applying it as:

$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \land x \in S \implies x \in B \land x \in T}$

The result follows directly from the definition of set intersection:

$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \cap S \implies x \in B \cap T}$

and from the definition of subset:

$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

$\blacksquare$