# Can we apply a successor operation in Peano's arithmetic infinitely many times, is the successor well defined?

If we look at the axioms of Peano arithmetic, e.g. http://mathworld.wolfram.com/PeanosAxioms.html, they contain an axiom:

If $a$ is a number, the successor of $a$ is a number.

However, the axioms do not limit how many times we could apply this successor operation. So we could apply successor to $0$ finitely many times, infinitely countably, uncountably, etc. So these axioms cannot define natural numbers because all the numbers generated would not be bijective with the natural numbers.

Is it a standard mathematical assumption that axioms can be applied only finitely many times to derive a sentence or construct a mathematical object? In set theory, there is an axiom of infinity, although we could construct any infinite set within infinite number of steps (applications if the other axioms).

When defining a successor, we apply it once and get a number $1$ from $0$. But how can we apply it once if we have only a definition of $0$ available? Is this not a problem?

• In ordinary logic, terms all have finite length. – André Nicolas Jan 19 '13 at 20:36

The axioms only allow finite iteration of $S$, just as you are only allowed finite iteration of logical axioms (infinite ones lead to a contradiction).

On the other hand there are strange "non-standard" models of PA which contains "infinitely" long applications of S: The reason for this is that first order logic is unable to uniquely define the notion of finiteness.

Everything you prove for arbitrary natural numbers by finitary means holds also for these non-standard numbers. It is not possible to write down a non-standard number though.

how can we apply it once if we have only a definition of 0 available?

there are two options

• sweep the issue under the rug: We work in some other metatheory which has already been defined
• break free: just do it - write S then write 0, no formal math needed use intuition and experience.