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. $\square$