It is known that the so-called continuum hypothesis \aleph_1=2^{\aleph_0}, where \aleph_1 (resp. \aleph_0) 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 \mathbb{R} with a cardinality \kappa satisfying \aleph_0 < \kappa < 2^{\aleph_0}.

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 \mathbb{R} (in fact of any uncountable Polish space) whose cardinality is strictly between \aleph_0 and 2^{\aleph_0}. Here, we generalize this result by showing that an uncountable borelian subset of \mathbb{R} (in fact of any Polish space) has cardinality greater than or equal to 2^{\aleph_0}.

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 2^{\aleph_0}.

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

Proof. Let Q \subset X be a countable dense subset, d : X \times X \to X be a distance compatible with the topology and \{U_i \mid i \in I\} be an open cover of X.

For all i \in I and q \in U_i \cap Q, let r_i(q)= \sup\{r >0 \mid B(q,r) \subset U_i\}>0.

Then \displaystyle U_i= \bigcup\limits_{q \in U_i \cap Q} B(q,r_i(q)) hence \displaystyle X= \bigcup\limits_{i \in I} \bigcup\limits_{q \in U_i \cap Q} B(q,r_i(q)).

For q \in Q fixed, let I(q)= \{ i_1,i_2, \dots \} \subset I be a countable subset so that (r_{i_n}(q)) be a sequence converging to \sup \{ r_i(q) \mid i \in I\} \in (0,+ \infty]. Therefore, for all i \in I there exists j\in I(q) such that r_j(q) \geq r_i(q).

Then \displaystyle X= \bigcup\limits_{q \in Q} \bigcup\limits_{i \in I(q)} B(q,r_i(q)).

For all (q,i) \in Q \times I(q) there exists k(q,i) \in I such that B(q,r_i(q)) \subset U_{k(q,i)}. Finally X= \bigcup\limits_{q \in Q} \bigcup\limits_{i \in I(q)} U_{k(q,i)} is a countable subcover. \square

Lemma 2: An uncountable Polish space has cardinality greater than or equal to 2^{\aleph_0}.

Proof. Let X be an uncountable Polish space and d : X \times X \to X be a complete distance compatible with the topology. Let C \subset X be the set of condensation points, that is points all of whose neighborhoods are uncountable. Then

  • C is not empty:

By contradiction, suppose that C= \emptyset. Then any point x \in X has a countable neighborhood B(x,r_x) (with r_x>0). According to lemma 1, \bigcup\limits_{x \in X} B(x,r_x) has a countable subcover. Then we deduce that X is countable as a countable union of countable sets, a contradiction.

  • C is closed:

Let (x_n) be a sequence in C converging to some x \in X and let \epsilon>0.  There exists p \in \mathbb{N} such that d(x_p,x)<\epsilon/2 hence B(x_p, \epsilon/2) \subset B(x,\epsilon). Because x_p \in C, B(x_p, \epsilon/2) is uncountable; we deduce that B(x,\epsilon) is also uncountable hence x \in C.

  • C has no isolated point.

Now, it is sufficient to prove that any complete metric space (Y,d) without isolated point is of cardinality at least 2^{\aleph_0}.

Let x_0,x_1 \in Y be two distinct points and r_0,r_1>0 be such that \overline{B}(x_0,r_0) \cap \overline{B}(x_1,r_1) = \emptyset. Then, let x_{00},x_{01} \in B(x_0,r_0) be two distinct points (B(x_0,r_0) contains at least two points since x_0 is not an isolated point) and r_{00},r_{01}>0 be such that B(x_{00},r_{00}),B(x_{01},r_{01}) \subset B(x_0,r_0) and \overline{B}(x_{00},r_{00}) \cap \overline{B}(x_{01},r_{01}) = \emptyset; we define in the same way x_{10}, x_{11}, r_{10}, r_{11}.

Thus, we construct by induction x_{\epsilon_1\dots \epsilon_n} for all n \geq 1 and \epsilon_1,\dots,\epsilon_n \in \{0,1\} such that

  • B(x_{\epsilon_1 \dots \epsilon_{n+1} },r_{\epsilon_1\dots \epsilon_{n+1}}) \subset B(x_{\epsilon_1\dots\epsilon_n}, r_{\epsilon_1 \dots \epsilon_n}),
  • \overline{B}(x_{\epsilon_1 \dots \epsilon_n}, r_{\epsilon_1 \dots \epsilon_n}) \cap \overline{B}(x_{\epsilon_1 \dots \overline{\epsilon_n}}, r_{\epsilon_1 \dots \overline{\epsilon_n}}) = \emptyset.

where \overline{\epsilon}=1- \epsilon for all \epsilon \in \{0,1\}.

Because the intersection of a decreasing sequence of closed sets cannot be empty in a complete metric space, there exists a point x(\epsilon_1, \epsilon_2,\dots) in \bigcap\limits_{n \geq 1} \overline{B}(x_{\epsilon_1 \dots \epsilon_n},r_{\epsilon_1\dots \epsilon_n}) for every (\epsilon_n) \in \{0,1\}^{\mathbb{N}}.

Finally, \left\{ \begin{array}{ccc} \{ 0,1 \}^{\mathbb{N}} & \to & Y \\ (\epsilon_n) & \mapsto & x(\epsilon_1, \epsilon_2, \dots) \end{array} \right. is an injective map, so the cardinality of Y is at least 2^{\aleph_0}. \square

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 2^{\aleph_0}, proving our theorem for closed subsets.

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

Proof. Let \{X_i \mid i \in I \subset \mathbb{N} \} be a countable family of Polish spaces. For each i \in I, let Q_i \subset X_i be a countable dense subset and d_i : X_i \times X_i \to X_i be a complete distance compatible with the topology; without loss of generality, we may suppose d_i \leq 1 (otherwise, replace it with the equivalent distance \min(d_i,1)).

First, \prod\limits_{i \in I} Q_i is a countable subspace of \prod\limits_{i \in I} X_i so \prod\limits_{i \in I} X_i is separable.

Then, introduce the distance \displaystyle \delta : (x,y) \mapsto \sum\limits_{i \in I} \frac{1}{2^i} d_i(x_i,y_i) on \prod\limits_{i \in I} X_i.

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

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

Proof. Let X_1,X_2 be two Polish spaces. For i=1,2 let Q_i \subset X_i be a countable dense subspace and d_i : X_i \times X_i \to X_i be a complete distance compatible with the topology; without loss of generality, we may suppose d_i \leq 1 (otherwise, replace it with the equivalent distance \min(d_i,1)).

First, Q_1 \oplus Q_2 is a countable dense subspace of X_1 \oplus X_2, so X_1 \oplus X_2 is separable.

Then, introduce the distance \delta : (x,y) \mapsto \left\{ \begin{array}{cl} d_1(x,y) & \text{if} \ x,y \in X_1 \\ d_2(x,y) & \text{if} \ x,y \in X_2 \\ 1 & \text{otherwise} \end{array} \right. on X_1 \oplus X_2.

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

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

Proof. Let X be a Polish space and O \subset X be an open subset.

Notice that \varphi : \left\{ \begin{array}{ccc} O & \to & X \times \mathbb{R} \\ x & \mapsto & \left( x, \frac{1}{d(x,X \backslash O)} \right) \end{array} \right. is a continuous injection.

Moreover, the projection (\pi_1)_{| \mathrm{Im}(\varphi)} : (x,r) \mapsto x is a continuous inverse of \varphi : O \to \mathrm{Im}(\varphi) , so \varphi is a homeomorphism onto \mathrm{Im}(\varphi).

To conclude that O is a Polish space, notice that \mathrm{Im}(\varphi) is closed in X \times \mathbb{R} and use lemma 3. Indeed, if (\varphi(x_n)) converges to (x,r), that is x_n \to x and \frac{1}{d(x_n,X \backslash O)} \to r, then d(x_n,X \backslash O) \to d(x,X \backslash O) so d(x,X \backslash O) \neq 0 and r= \frac{1}{d(x,X \backslash O)}. \square

Lemma 6: Let (X, \mathcal{T}) be a Polish space and B \subset X be a borelian subset. Then there exists a topology \mathcal{T}_B \supset \mathcal{T} on X such that (X,\mathcal{T}_B) is a Polish space and B is closed in (X,\mathcal{T}_B).

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

  • Closed subsets belong to \mathfrak{B}.

If B \subset X is closed, then just take \mathcal{T}_B= \mathcal{T}.

  • \mathfrak{B} is stable under complement.

Let B \in \mathfrak{B} and let \mathcal{T}_{X \backslash B} be the topology generated by \mathcal{T}_{B} \cup \{B\}. To conclude, notice that (X,\mathcal{T}_{X \backslash B}) is in fact homeomorphic to the direct sum B \oplus (X \backslash B) and use lemmas 4 and 5.

  • \mathfrak{B} is stable under countable union.

Let B_n \in \mathfrak{B} be a countable family and \mathcal{T}_n be the corresponding topologies. Let \mathcal{T}_{\infty} be the topology generated by \bigcup\limits_{n \geq 1} \mathcal{T}_n.

Notice that \varphi : \left\{\begin{array}{ccc} (X, \mathcal{T}_{\infty}) & \to & \prod\limits_{n \geq 1} (X,\mathcal{T}_n) \\ x & \mapsto & (x,x,\dots) \end{array} \right. is a continuous injection.

Moreover, if \{ O_n \in \mathcal{T}_n \mid n \in K\} is a finite collection of open sets, then

\varphi \left( \bigcup\limits_{n \in k} O_n \right) = (\Omega_1 \times \Omega_2 \times \cdots ) \cap \mathrm{Im}(\varphi) where \Omega_i= \left\{ \begin{array}{cl} O_i & \text{if} \ i \in K \\ X & \text{otherwise} \end{array} \right.

Therefore, \varphi is a homeomorphism onto \mathrm{Im}(\varphi). Then, it is sufficient to show that \mathrm{Im}(\varphi) is closed in \prod\limits_{n \geq 0} (X,\mathcal{T}_n) to conlude that (X,\mathcal{T}_{\infty}) is a Polish space since \prod\limits_{n \geq 1} (X,\mathcal{T}_n) is itself a Polish space according to lemma 3.

Let (x_n) \in \left( \prod\limits_{n \geq 1} X \right) \backslash \mathrm{Im}(\varphi). Then there exist i \neq j such that x_i \neq x_j, and because (X, \mathcal{T}) is Hausdorff there exist U,V \in \mathcal{T} separating x_i and x_j. Let

\Omega= \prod\limits_{n \geq 1} \Omega_n where \Omega_n= \left\{ \begin{array}{cl} U & \text{if} \ n=i \\ V & \text{if} \ n=j \\ X & \text{otherwise} \end{array} \right..

Because \mathcal{T} \subset \bigcap\limits_{n \geq 1} \mathcal{T}_n, \Omega \in \mathcal{T}_{\infty}. Moreover, (x_n) \in \Omega and \Omega \subset \left( \prod\limits_{n \geq 1} X \right) \backslash \mathrm{Im}(\varphi) since U \cap V = \emptyset. \square

Proof of the theorem: Let (X,\mathcal{T}) be a Polish space and B \subset X be a borelian subset. Let \mathcal{T}_B denote the topology given by the lemma 6. Then B can be viewed as a Polish space as a closed subspace of (X, \mathcal{T}_B). From lemma 2, we deduce that B is of cardinality at least 2^{\aleph_0} if it is uncountable. \square

Advertisements