Cardinality of Proper Subset of Finite Set

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

Let $\card B = n$, where $\card {\, \cdot \,}$ denotes cardinality.

Then $\card A < n$.

Proof
The proof proceeds by the Principle of Mathematical Induction on $n$, the cardinality of $B$.

Let $S$ be the set of $n \in \N$ such that every proper subset of any set with $m$ elements is finite and has (strictly) fewer than $n$ elements.

Basis for the Induction
The case $n = 0$ is verified as follows:

Suppose $\card B = 0$.

From Cardinality of Empty Set, $B = \O$.

There are no sets $A$ such that $A \subsetneq \O$.

Therefore $0 \in S$ by vacuous truth.

This is the basis for the induction.

Induction Hypothesis
Fix $k \in \N$ with $k \ge 0$.

Suppose that $k \in S$.

That is:
 * every proper subset of any set with $k$ elements is finite and has (strictly) fewer than $k$ elements.

This is our induction hypothesis.

It remains to be proved that $k + 1 \in S$:
 * every proper subset of any set with $k + 1$ elements is finite and has (strictly) fewer than $k + 1$ elements.

Induction Step
This is our induction step:

Suppose we have any set $B$ such that $\card B = k + 1$.

Let $A \subsetneqq B$.

From the definition of $\subsetneqq$:


 * $\exists b \in B: b \notin A$

Therefore:
 * $A \subseteq B \setminus \set b$

From Cardinality Less One:
 * $\card {B - \set b} = k$

Suppose $A = B - \set b$.

Then $\card A = k < k + 1$ and so has fewer than $k + 1$ elements and is finite by definition.

Otherwise:
 * $A \subsetneqq B - \set b$

So by the induction hypothesis:
 * $A$ is finite and has fewer than $k$ elements.

But as $k < k + 1$ it also follows that $A$ is finite and has fewer than $k + 1$ elements.

Thus $A \in S$.

The result follows by the Principle of Mathematical Induction.