Let be an uncountable well-ordered set, and let stand for the minimum element of . We denote by the set of all elements such that the set of its predecessors is uncountable. If the set is empty, we will replace by , where , and set for all .

Since is a nonempty subset of the well-order set , there exists a unique minimum element called first uncountable ordinal. Therefore, if then is countable.

**Exercise**. Give an example for an uncountable set with a well-order relation.

**Theorem 1.** (Transfinite induction)* Let be a property defined for all . Suppose that:*

*1. is true,*

*2. for each , if is true for all , then is also true.*

*Therefore, is true for all .*

Proof.

Let be the set of such that is false. Suppose that . Since is nonempty subset of the well-ordered set , there exists an unique minimum element . Note that and the set of its predecessors is countable. Hence, the property is true for 1 and all predecessors of . It follows that is also true. This inconsistency completes our proof.

**Theorem 2.** *For any sequence in , , there exists such that .*

Proof. Note that is countable. Therefore, we can choose as an element in the uncountable set .

Now let stand for the family of all open sets, and be the family of all closed sets (considered in a topology space). Suppose that , are defined for all , then we can define

By the principle of transfinite induction, are well-defined for every .

**Lemma 3.** *1. for *

*and for then*

*2. , ;*

*3. , .*

Proof.

1. The fact is true for . Assume that it is true for all , then

for each , we have , where . Moreover, . It allows that . Therefore, by the principle of transfinite induction, we conclude that and similarly, for .

2. This fact is obvious.

3. If , then where . This immediately follows that . Thus and similarly, .

**Theorem 4. ***The Borel sigma-algebra, i.e. smallest sigma-algebra that consists all open (closed) sets, is given by *

Proof. Lemma 3.2 help us to prove the identity .

It is easy to show that is closed under complement by applying Lemma 3.1.

If then .

Applying Theorem 2, there exists such that . It follows that

Moreover, one can show by the transfinite induction that are subsets of any sigma algebra that contains all closed (open) sets. Hence, we can conclude that is the Borel sigma-algebra that we want to understand its structure.