Set is Subset of Intersection of Supersets
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Theorem
Let $S$, $T_1$ and $T_2$ be sets.
Let $S$ be a subset of both $T_1$ and $T_2$.
Then:
- $S \subseteq T_1 \cap T_2$
That is:
- $\paren {S \subseteq T_1} \land \paren {S \subseteq T_2} \implies S \subseteq \paren {T_1 \cap T_2}$
Set of Sets
Let $T$ be a set.
Let $\mathbb S$ be a set of sets.
Suppose that for each $S \in \mathbb S$, $T \subseteq S$.
Then:
- $T \subseteq \ds \bigcap \mathbb S$
General Result
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $X$ be a set such that:
- $\forall i \in I: X \subseteq S_i$
Then:
- $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
Proof 1
Let $S \subseteq T_1 \land S \subseteq T_2$.
Then:
\(\ds x \in S\) | \(\leadsto\) | \(\ds x \in T_1 \land x \in T_2\) | Definition of Subset | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x \in T_1 \cap T_2\) | Definition of Set Intersection | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds S \subseteq T_1 \cap T_2\) | Definition of Subset |
Proof 2
\(\ds S\) | \(\subseteq\) | \(\ds T_1\) | ||||||||||||
\(\, \ds \land \, \) | \(\ds S\) | \(\subseteq\) | \(\ds T_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds S \cap S\) | \(\subseteq\) | \(\ds S \cap T_2\) | Set Intersection Preserves Subsets | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds T_1 \cap T_2\) | Set Intersection is Idempotent |
$\blacksquare$
Also see
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{B viii}$
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(iv)}$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.6$: Set Identities and Other Set Relations: Exercise $4$