It is known that the so-called continuum hypothesis , where (resp. ) is the cardinality of the first uncountable ordinal (resp. the first infinite ordinal), is indecidable in ZFC. In other words, we cannot know (in ZFC) wether there exists a subset of with a cardinality satisfying .

However, it is possible to answer the question by restricting the class of sets studied. For example, Cantor-Bendixson theorem (see the remark following lemma 2) can be used to prove that there is no closed subset of (in fact of any uncountable Polish space) whose cardinality is strictly between and . Here, we generalize this result by showing that an uncountable borelian subset of (in fact of any Polish space) has cardinality greater than or equal to .

More information about borelian subsets of a Polish space can be found in Srivastava’s book, *A Course on Borel Sets*. In particular, a more general version of our theorem is proved: An uncoutable borelian subset of a Polish space contains a subspace homeomorphic to the Cantor set.

**Theorem:** An uncountable borelian subset of a Polish space has cardinality greater than or equal to .

**Lemma 1:** Let be a metrizable space. If is separable, then is a Lindelöf space.

**Proof.** Let be a countable dense subset, be a distance compatible with the topology and be an open cover of .

For all and , let .

Then hence .

For fixed, let be a countable subset so that be a sequence converging to . Therefore, for all there exists such that .

Then .

For all there exists such that . Finally is a countable subcover.

**Lemma 2:** An uncountable Polish space has cardinality greater than or equal to .

**Proof.** Let be an uncountable Polish space and be a complete distance compatible with the topology. Let be the set of condensation points, that is points all of whose neighborhoods are uncountable. Then

- is not empty:

By contradiction, suppose that . Then any point has a countable neighborhood (with ). According to lemma 1, has a countable subcover. Then we deduce that is countable as a countable union of countable sets, a contradiction.

- is closed:

Let be a sequence in converging to some and let . There exists such that hence . Because , is uncountable; we deduce that is also uncountable hence .

- has no isolated point.

Now, it is sufficient to prove that any complete metric space without isolated point is of cardinality at least .

Let be two distinct points and be such that . Then, let be two distinct points ( contains at least two points since is not an isolated point) and be such that and ; we define in the same way .

Thus, we construct by induction for all and such that

- ,
- .

where for all .

Because the intersection of a decreasing sequence of closed sets cannot be empty in a complete metric space, there exists a point in for every .

Finally, is an injective map, so the cardinality of is at least .

**Remark:** The first part of the proof can be used to prove Cantor-Bendixson theorem, stating that a closed subset of a Polish space can be (uniquely) written as the disjoint union of a perfect subset and a countable subset; useful material can be found in our previous post on Cantor-Bendixson rank. The second part of the proof then shows that the perfect subset has to be of cardinality at least , proving our theorem for closed subsets.

**Lemma 3:** The direct product of countably many Polish spaces is a Polish space.

**Proof.** Let be a countable family of Polish spaces. For each , let be a countable dense subset and be a complete distance compatible with the topology; without loss of generality, we may suppose (otherwise, replace it with the equivalent distance ).

First, is a countable subspace of so is separable.

Then, introduce the distance on .

It is a classical exercice to prove that is compatible with the product topology and complete. The details are left to the reader.

**Lemma 4:** The direct sum of two Polish spaces is a Polish space.

**Proof.** Let be two Polish spaces. For let be a countable dense subspace and be a complete distance compatible with the topology; without loss of generality, we may suppose (otherwise, replace it with the equivalent distance ).

First, is a countable dense subspace of , so is separable.

Then, introduce the distance on .

Finally, it is easy to show that is compatible with the topology and complete.

**Lemma 5:** An open subspace of a Polish space is a Polish space.

**Proof.** Let be a Polish space and be an open subset.

Notice that is a continuous injection.

Moreover, the projection is a continuous inverse of , so is a homeomorphism onto .

To conclude that is a Polish space, notice that is closed in and use lemma 3. Indeed, if converges to , that is and , then so and .

**Lemma 6:** Let be a Polish space and be a borelian subset. Then there exists a topology on such that is a Polish space and is closed in .

**Proof.** Let be the set of subsets satisfying the property. Then, to show that borelian sets are included in , it is sufficient to notice that contains the closed subsets and is stable under complement and countable union.

- Closed subsets belong to .

If is closed, then just take .

- is stable under complement.

Let and let be the topology generated by . To conclude, notice that is in fact homeomorphic to the direct sum and use lemmas 4 and 5.

- is stable under countable union.

Let be a countable family and be the corresponding topologies. Let be the topology generated by .

Notice that is a continuous injection.

Moreover, if is a finite collection of open sets, then

where

Therefore, is a homeomorphism onto . Then, it is sufficient to show that is closed in to conlude that is a Polish space since is itself a Polish space according to lemma 3.

Let . Then there exist such that , and because is Hausdorff there exist separating and . Let

where .

Because , . Moreover, and since .

**Proof of the theorem:** Let be a Polish space and be a borelian subset. Let denote the topology given by the lemma 6. Then can be viewed as a Polish space as a closed subspace of . From lemma 2, we deduce that is of cardinality at least if it is uncountable.