Union with Empty Set/Proof 1

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

$\blacksquare$

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