Inductive Sets: Why a Table of Cases Is Not a Proof

Tabulate $5^n - 1$ for $n = 1$ to $6$ as multiples of 4, or check $1^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ for several $n$, and state that no finite table proves $(\forall n \in \mathbb{N})(P(n))$.

Write the definition as $(\forall k \in \mathbb{Z})(k \in T \to k + 1 \in T)$, and record that $\mathbb{N}$, $\{n \in \mathbb{Z} \mid n \geq -3\}$ and $\mathbb{Z}$ are all inductive despite starting in different places or nowhere.

Negate the definition step by step to reach $(\exists k \in \mathbb{Z})(k \in T \wedge k + 1 \notin T)$, and give the single counterexample for $\{1, 2, \ldots, 20\}$, $\{n \in \mathbb{Z} \mid n \leq 100\}$ and the odd naturals.

Starting from $T \subseteq \mathbb{N}$, $1 \in T$ and $T$ inductive, derive $2, 3, 4 \in T$ in turn, describe reaching $100 \in T$, and note that concluding $T = \mathbb{N}$ rests on an unproved 'and so on'.