## Theorem 2.3.3: Induction Principle

Let

**S**be a well-ordered set with the additional property that every element except for the smallest one has an immediate predecessor. Then: if**Q**is a property such that:- the smallest element of
**S**has the property**Q** - if
*s*has property**S****Q**then the successor of*s*also has property**Q**

**Q**holds for every element in**S**.### Proof:

It is not quite obvious what it is that we have to prove: we need to show that if a property holds for the smallest element of a well-ordered set**S**, and it holds for every successor of a general element of that set, then it is indeed true for the whole set

**S**.

We will prove this by contradiction: Suppose property **Q** is true
for the smallest element of a well-ordered set **S**, denoted by the
symbol 1. Suppose also that property **Q** is true for the successor
*n + 1* of the general element *n* of that set, if it is assumed
to be true for that element *n*.

- Denote by
**E**the set of all elements from**S**for which property**Q**is not true

**S**is well-ordered, the subset

**E**contains a smallest element, say

*n*. If

*n = 1*, we would have a contradiction immediately. If

*n > 1*, then it must have an immediate predecessor, denoted by the element

*n-1*.

For that element the property **Q** holds true, and therefore, being true
for *n - 1* it must be true for *(n-1) + 1 = n*, by assumption.
Hence, the set **E** must be empty, and property **Q** is therefore
true for all of **S**.