Intersection Operation on Supersets of Subset is Closed
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Theorem
Let $S$ be a set.
Let $T \subseteq S$ be a given subset of $S$.
Let $\powerset S$ denote the power set of $S$
Let $\mathscr S$ be the subset of $\powerset S$ defined as:
- $\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
Then the algebraic structure $\struct {\mathscr S, \cap}$ is closed.
Proof
Let $A, B \in \mathscr S$.
We have that:
\(\ds T\) | \(\subseteq\) | \(\ds A\) | Definition of $\mathscr S$ | |||||||||||
\(\ds T\) | \(\subseteq\) | \(\ds B\) | Definition of $\mathscr S$ | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds T\) | \(\subseteq\) | \(\ds A \cap B\) | Intersection is Largest Subset |
and:
\(\ds A\) | \(\subseteq\) | \(\ds S\) | Definition of Power Set | |||||||||||
\(\ds B\) | \(\subseteq\) | \(\ds S\) | Definition of Power Set | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap B\) | \(\subseteq\) | \(\ds S\) | Intersection is Subset | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds A \cap B\) | \(\in\) | \(\ds \powerset S\) | Definition of Power Set |
Thus we have:
\(\ds T\) | \(\subseteq\) | \(\ds A \cap B\) | from $(1)$ | |||||||||||
\(\ds A \cap B\) | \(\in\) | \(\ds \powerset S\) | from $(2)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap B\) | \(\in\) | \(\ds \mathscr S\) | Definition of $\mathscr S$ |
Hence the result by definition of closed algebraic structure.
$\blacksquare$
Also see
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $5$: Subgroups: Exercise $1$