Union with Superset is Superset/Proof 1
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Theorem
- $S \subseteq T \iff S \cup T = T$
Proof
Let $S \cup T = T$.
Then by definition of set equality:
- $S \cup T \subseteq T$
Thus:
\(\ds S\) | \(\subseteq\) | \(\ds S \cup T\) | Subset of Union | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds T\) | Subset Relation is Transitive |
Now let $S \subseteq T$.
From Subset of Union, we have $S \cup T \supseteq T$.
We also have:
\(\ds S\) | \(\subseteq\) | \(\ds T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds S \cup T\) | \(\subseteq\) | \(\ds T \cup T\) | Set Union Preserves Subsets | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S \cup T\) | \(\subseteq\) | \(\ds T\) | Set Union is Idempotent |
Then:
\(\ds S \cup T\) | \(\subseteq\) | \(\ds T\) | ||||||||||||
\(\ds S \cup T\) | \(\supseteq\) | \(\ds T\) |
By definition of set equality:
- $S \cup T = T$
So:
\(\ds S \cup T = T\) | \(\implies\) | \(\ds S \subseteq T\) | ||||||||||||
\(\ds S \subseteq T\) | \(\implies\) | \(\ds S \cup T = T\) |
and so:
- $S \subseteq T \iff S \cup T = T$
from the definition of equivalence.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(f)}$
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- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): Appendix $\text{A}.2$: Theorem $\text{A}.11$