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:
where for all ,
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. :
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.