Union with Superset is Superset

Theorem

$S \subseteq T \iff S \cup T = T$

where:

• $S \subseteq T$ denotes that $S$ is a subset of $T$;
• $S \cup T$ denotes the union of $S$ and $T$.

Proof

Let $S \cup T = T$.

Then from the definition of set equality, $S \cup T \subseteq T$.

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S\)	\(\subseteq\)	\(\displaystyle S \cup T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subset of Union
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subsets Transitive

---

Now let $S \subseteq T$.

From Subset of Union, we have $S \cup T \supseteq T$.

We also have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S \cup T\)	\(\subseteq\)	\(\displaystyle T \cup T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Union Preserves Subsets
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S \cup T\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Union is Idempotent

So as we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \cup T\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \cup T\)	\(\supseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

it follows from the definition of Set Equality that we have $S \cup T = T$.

---

So we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \cup T = T\)	\(\implies\)	\(\displaystyle S \subseteq T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S \subseteq T\)	\(\implies\)	\(\displaystyle S \cup T = T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

and so:

$S \subseteq T \iff S \cup T = T$

from the definition of equivalence.

$\blacksquare$

Also see

• Intersection with Subset is Subset

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 4$: Unions and Intersections
• W.E. Deskins: Abstract Algebra (1964): $\S 1.1$: Exercise $1.1: \ 9$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $3.3 \ \text{(d)}$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Exercises $\text{B iv}$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 6 \ \text{(d)}$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 7.3 \ \text{(i)}$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.2$: Exercise $1.2.1 \ \text{(vii)}$