Principle of transfinite induction for ordinals:
Let $A$ be a subset of the ordinal $\alpha$ such that
- $0 \in A$
- for all $\beta \in \alpha$ if $\beta \subseteq A$ then $\beta \in A$
Then $A=\alpha$
Isn't the first condition $0\in A$ implicit in the second condition?
"For all $\beta \in \alpha$ if $\beta \subseteq A$ then $\beta \in A$" means that, for all the elements $\beta$ in $\alpha$ (therefore $0$ included) this holds: $(\delta \in \beta$ ⇒ $\delta \in A) ⇒ \beta \in A$ the antecedent of this is vacuously true for $\beta=0$ and therefore if this property has to hold that means that $0\in A$ which is just the first condition. Isn't this correct? Is the condition $0 \in A$ there just to help avoid mistakes or it's actually needed?
The first condition is not implicit in the second: If $\alpha=0$, then clearly $0\notin A$ because $0\notin\alpha=0$, but the second condition is vacuously fulfilled. Therefore $\alpha=0$ is a counterexample to the claim that the second condition implies the first.
However it is the only counterexample, and for $\alpha=0$, $A=\alpha$ trivially holds. Thus the theorem obtained by removing the first condition also is true.
Therefore the first condition, while not implicit in the second, is still unnecessary.