Cardinality of Subset of Finite Set

Theorem
Let $A$ and $B$ be finite sets such that $A \subseteq B$.

Let $\left|{B}\right| = n$.

Then $\left|{A}\right| \le n$.

If $A \ne B$, i.e. $A \subset B$, then $\left|{A}\right| < n$.

Proof
Let $S$ be the set of all elements $m \in \N$ such that:

$\forall B: \left|{B}\right| = m, A \subset B: \left|{A}\right| < m$


 * From Cardinality of Empty Set, $\left|{\varnothing}\right| = 0$. There are also no sets $T$ such that $T \subset \varnothing$, so $0 \in S$.


 * Let $m \in S$.

Let $\left|{B}\right| = m + 1$.

Let $A \subset B$, i.e. $A \subseteq B$ and $A \ne B$.

Then $\exists b \in B: b \notin A$.

Thus $A \subseteq \left({B - \left\{{b}\right\}}\right)$.

By Cardinality Less One, $\left|{B - \left\{{b}\right\}}\right| = m$.

There are two possibilities:


 * 1) If $A = B - \left\{{b}\right\}$, then $\left|{A}\right| = m$;
 * 2) If $A \subset \left({B - \left\{{b}\right\}}\right)$, then $\left|{A}\right| < m$, since $m \in S$.

Since $m < m + 1$, it follows that $\left|{A}\right| < m + 1$.

Hence $m + 1 \in S$.


 * By induction, $S = \N$.

In particular, $n \in S$.