Set Intersection Preserves Subsets

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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:

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


Proof

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

Then:

\(\displaystyle x \in A\) \(\implies\) \(\displaystyle x \in B\) Definition of Subset
\(\displaystyle x \in S\) \(\implies\) \(\displaystyle x \in T\) Definition of Subset


Now we invoke the Praeclarum Theorema of propositional logic:

$\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$

applying it as:

$\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \land x \in S \implies x \in B \land x \in T}\right)$

The result follows directly from the definition of set intersection:

$\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \cap S \implies x \in B \cap T}\right)$

and from the definition of subset:

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

$\blacksquare$