Here we prove that an uncountable separable metric space has very few Borel subsets. More precisely:

**Theorem:** Let be an infinite separable metric space and let denote the set of Borel subsets of . Then .

As a corollary, we deduce that has subsets whose only are Borel sets.

First, we need to define the Borel hierarchy:

**Definition:** Let be a metric space and let (resp. ) denote the family of open (resp. closed) subsets of . Then, we define and for any ordinal by transfinite induction using:

and ,

where for all ,

and .

Finally, let for all .

**Lemma1:** Let be an ordinal and . Then if and only if .

**Proof.** Easy by transfinite induction. The details are left to the reader.

**Lemma 2:** Let be a metric space. We have the following inclusions:

**Proof.** First, the inclusions are clear.

Let . Clearly, is a -set ie. . Moreover, it is a -set ie. :

where .

Therefore, . Using lemma 1, we also deduce .

Now, it is easy to show that by transfinite induction.

**Lemma 3:** Let be a metric space. Then .

**Proof.** It is easy to show by transfinite induction that for all , hence

.

For convenience, let . Then

- contains the open subset of .
- is stable under countable union:

Let . For all , be an ordinal such that . Let . Using lemma 2,

.

- is stable under complement:

Let and such that . According to lemmas 1 and 2,

.

Therefore, and finally .

The other equalities follows from given by lemma 2.

**Proof of the theorem:** First, we show that .

Let be a countable dense subset and be an open subspace. For all , define ; if , let . For fixed, let be an increasing sequence of rationals converging to . Then,

.

The map defines an injection , hence . In particular, we also have .

Because for all , it is easy to show by transfinite induction that for all .

Using lemma 3, we deduce that

.

Moreover, so .

For more information, we can see Srivastava’s book, *A Course on Borel Sets*. In particular, a topological proof of our theorem for Polish spaces can be found there.