Union with Empty Set/Proof 1
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Theorem
The union of any set with the empty set is the set itself:
- $S \cup \O = S$
Proof
| \(\ds S\) | \(\subseteq\) | \(\ds S\) | Set is Subset of Itself | |||||||||||
| \(\ds \O\) | \(\subseteq\) | \(\ds S\) | Empty Set is Subset of All Sets | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds S \cup \O\) | \(\subseteq\) | \(\ds S\) | Union is Smallest Superset | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds S \cup \O\) | Set is Subset of Union | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds S \cup \O\) | \(=\) | \(\ds S\) | Definition 2 of Set Equality |
$\blacksquare$