Intersection Largest

Theorem

Let $T_1$ and $T_2$ be sets.

Then $T_1 \cap T_2$ is the largest set contained in both $T_1$ and $T_2$.

That is:

$S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$

General Result

Let $T$ be a set.

Let $\mathcal P \left({T}\right)$ be the power set of $T$.

Let $\mathbb T$ be a subset of $\mathcal P \left({T}\right)$.

Then:

$\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$

Proof

• Let $S \subseteq T_1 \land S \subseteq T_2$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \in S\)	\(\implies\)	\(\displaystyle x \in T_1 \land x \in T_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Subset
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle x \in T_1 \cap T_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Intersection
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle S \subseteq T_1 \cap T_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Subset

Alternatively:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S\)	\(\subseteq\)	\(\displaystyle T_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S\)	\(=\)	\(\displaystyle S \cap T_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Intersection with Subset is Subset‎
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\subseteq\)	\(\displaystyle T_1 \cap T_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Intersection Preserves Subsets

So:

$S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$.

• Now let $S \subseteq T_1 \cap T_2$.

From Intersection Subset we have $T_1 \cap T_2 \subseteq T_1$ and $T_1 \cap T_2\subseteq T_2$.

From Subsets Transitive, it follows directly that $S \subseteq T_1$ and $S \subseteq T_2$.

So $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.

• From the above, we have:

1. $S \subseteq T_1 \land S \subseteq T_2 \implies S \subseteq T_1 \cap T_2$;
2. $S \subseteq T_1 \cap T_2 \implies S \subseteq T_1 \land S \subseteq T_2$.

Thus $S \subseteq T_1 \land S \subseteq T_2 \iff S \subseteq T_1 \cap T_2$ from the definition of equivalence.

$\blacksquare$

Proof of General Result

Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.

Suppose that $\forall X \in \mathbb T: S \subseteq X$.

Consider any $x \in S$.

By definition of subset, it follows that:

$\forall X \in \mathbb T: x \in X$

Thus it follows from definition of set intersection that:

$x \in \bigcap \mathbb T$

Thus by definition of subset, it follows that:

$S \subseteq \bigcap \mathbb T$

So:

$\left({\forall X \in \mathbb T: S \subseteq X}\right) \implies S \subseteq \bigcap \mathbb T$

$\Box$

Now suppose that $S \subseteq \bigcap \mathbb T$.

From Intersection Subset: General Result we have:

$\forall X \in \mathbb T: \bigcap \mathbb T \subseteq X$

So from Subsets Transitive, it follows that:

$\forall X \in \mathbb T: S \subseteq \bigcap \mathbb T \subseteq X$

So it follows that $\forall X \in \mathbb T: S \subseteq X$.

So:

$S \subseteq \bigcap \mathbb T \implies \left({\forall X \in \mathbb T: S \subseteq X}\right)$

$\Box$

Hence:

$\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$

$\blacksquare$

Caution

Be careful of the way the general result is stated:

$\left({\forall X \in \mathbb T: S \subseteq X}\right) \iff S \subseteq \bigcap \mathbb T$

Without the brackets, this would read:

$\forall X \in \mathbb T: S \subseteq X \iff S \subseteq \bigcap \mathbb T$

That is:

For all $X \in \mathbb T$, it is the case that $S \subseteq X$ if and only if $S \subseteq \bigcap \mathbb T$

This is not the same thing at all.

For example, if $\mathbb T = \mathcal P \left({T}\right)$ (as it well might), then $T \in \mathbb T$, and $\bigcap \mathbb T = \varnothing$.

This would imply that:

$T \subseteq \bigcap \mathbb T$

That is:

$T \subseteq \varnothing$

which is obviously rubbish.

Also see

• Union Smallest

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 1.3$: Example $13$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $3.6 \ \text{(e)}$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Exercises $\text{B viii}$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1, \S 6$, Theorem $6.1 \ (1)$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 7.2 \ \text{(i)}$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.2$: Exercise $1.2.1 \ \text{(iv)}$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.4$: Exercise $1.4.4 \ \text{(ii)}$