\section{Set Intersection Preserves Subsets/Corollary/Proof 1}
Tags: Set Intersection Preserves Subsets

\begin{theorem}
Let $A, B, S$ be [[Definition:Set|sets]].
Then:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
\end{theorem}

\begin{proof}
Let $A \subseteq B$, and let $S$ be any set.
From [[Set Intersection Preserves Subsets]]:
:$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
:$A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From [[Set is Subset of Itself]], $S \subseteq S$ for all sets $S$.
Hence the result:
:$A \subseteq B \implies A \cap S \subseteq B \cap S$
{{qed}}
\end{proof}


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