Prove that the "number of elements" set-function satisfies Axioms 1-3 of area

Let

$A = \{ 1,2,3,4,5 \}, \qquad \mathcal{M} = \mathcal{P}(A),$

where $\mathcal{P}(A)$ is power set of $A$ (i.e., the class of all subsets of $A$ ). Next, define a function $n$ by
$n(S)$ equals the number of distinct elements in $S$ for any $S \in \mathcal{M}$ . Then, let

$S = \{ 1,2,3,4 \}, \qquad T = \{ 3,4,5 \}.$

Compute

$n(S \cup T), \quad n(S \cap T), \quad n(S - T), \quad n(T-S).$

Then, prove the function $n$ satisfies Axioms 1-3 for area.

First, we compute

$\begin{alignat*}{2} n(S \cup T) &= n( \{ 1,2,3,4,5 \}) &= 5 \\ n(S \cap T) &= n( \{ 3,4 \} ) &= 2 \\ n(S - T) &= n(\{1,2 \}) &= 2 \\ n(T - S) &= n(\{ 5 \}) &= 1 \end{alignat*}\end{align*}$

Next, we prove this satisfies the first three area axioms.
Proof.
Axiom 1. (Non-negative property) This is satisfied for any set $S$ since the number of distinct elements in a set is non-negative. So, $n(S) \geq 0$ for all $S$ .
Axiom 2. (Additive property) First, if $S,T \in \mathcal{M}$ , then $S \subseteq A, \ T \subseteq A$ by definition of $\mathcal{M}$ . So, for any $x \in S$ we have $x \in A$ and for any $y \in T$ , we have $y \in A$ .
Thus, if $x \in S \cup T$ , then $x \in A$ ; hence, $S \cup T \subseteq A$ , so $S \cup T \in \mathcal{M}$ .
Then, $S \cap T \subseteq S$ implies $S \cap T \subseteq A$ (since $S \subseteq A$ ). Hence, $S \cap T \in \mathcal{M}$ .
So, for any $S,T \in \mathcal{M}$ we have $S \cup T \in \mathcal{M}, \ S \cap T \in \mathcal{M}$ .
Next, we must show $n(S \cup T) = n(S) + n(T) - n(S \cap T)$ . For any $x \in S \cup T$ we have $x \in S, \ x \in T$ , or $x \in S$ and $T$ . So, this means $x \in (S - T)$ , or $x \in (T - S)$ or $x \in (S \cap T)$ . Thus,

$n(S \cup T) = n(S-T) + n(T-S) + n(S \cap T).$

Similarly, we note,

$\begin{align*} n(S) = n(S-T) + n(S \cap T) \quad &\implies \quad n(S-T) = n(S) - n(S \cap T) \\ n(T) = n(T-S) + n (T \cap S) \quad &\implies \quad n(T-S) = n(T) - n(S \cap T) \end{align*}$