Tests for Finite Set Equality
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Theorem
Let $S$ and $T$ be finite sets.
Let $S \subseteq T$.
The following statements are equivalent:
- $(a): \quad \card S = \card T$
- $(b): \quad \card {T \setminus S} = 0$
- $(c): \quad T \setminus S = \O$
- $(d): \quad T = S$
Proof
We have:
\(\text {(a)}: \quad\) | \(\ds \card S\) | \(=\) | \(\ds \card T\) | |||||||||||
\(\text {(b)}: \quad\) | \(\ds \leadstoandfrom \ \ \) | \(\ds \card {T \setminus S}\) | \(=\) | \(\ds \card T - \card S = 0\) | Cardinality of Set Difference with Subset | |||||||||
\(\text {(c)}: \quad\) | \(\ds \leadstoandfrom \ \ \) | \(\ds T \setminus S\) | \(=\) | \(\ds \O\) | Cardinality of Empty Set | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds T\) | \(\subseteq\) | \(\ds S\) | Set Difference with Superset is Empty Set | ||||||||||
\(\text {(d)}: \quad\) | \(\ds \leadstoandfrom \ \ \) | \(\ds T\) | \(=\) | \(\ds S\) | Definition of Set Equality |