Principle of Recursive Definition

Theorem
Let $\N$ be the natural numbers.

Let $T$ be a class (which may be a set).

Let $a \in T$.

Let $g: T \to T$ be a mapping.

Then there exists exactly one mapping $f: \N \to T$ such that:


 * $\forall x \in \N: \map f x = \begin{cases}

a & : x = 0 \\ \map g {\map f n} & : x = n + 1 \end{cases}$

Fallacious Proof
The proofs given (hidden behind the links) are necessarily long, precise and detailed.

There is a temptation to take short cuts and gloss over the important details.

The following argument, for example, though considerably shorter, is incorrect.

Also known as
This result is often referred to as the Recursion Theorem.

Some sources only cover the general result.

Also see

 * Second Principle of Recursive Definition


 * Transfinite Recursion Theorem