Cardinality of Subset of Finite Set

Theorem
Let $$A$$ and $$B$$ be 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, $$n \in S$$.