Can every dense subset be partitioned into two dense subsets?

Question

The question was very popular over on MSE but seems to have left everyone speechless. Maybe someone here can help?

Definition: Suppose $X$ is a compact connected Hausdorff space and $D \subset X$ countable and dense. We say $D$ is divisible to mean for every open $U \subset X$ there exists a partition $D \cap U = D_1 \cup D_2$ with $D_1$ and $D_2$ both dense in $U$. We call $X$ divisible to mean it is separable and each countable dense subset is divisible.

Is every compact connected Hausdorff space divisible?

The property is easy to show if $X$ is metric rather than just Hausdorff. Just let $U_1,U_2, \ldots $ be a countable basis. Since $X$ is connected each $U_n$ contains a proper subcontinuum which surjects onto $[0,1]$ by Urysohn's lemma. Hence each open subset is uncountable. So choose distinct $a_1,b_1 \in U_1$ and proceed by induction. At each stage we have selected only finitely many $a_n,b_n$ and $U_m$ contains infinitely many other elements. So select distinct $a_m,b_m \in U_m - \{a_1,b_2, \ldots , a_{m-1},b_{m-1}\}$ and define $D_1 = \{a_1,a_2,\ldots\}$ and $D_2 = \{b_1,b_2,\ldots\} \cup (D - D_1)$. Then $D_1 \cap U$ and $D_2 \cap U$ form the desired partition for any choice of $U$. So $X$ might be called simultaneously divisible.

The property fails for (disconnected) spaces with an isolated point. Because if $\{x\} \subset X$ is open $x$ must be an element of exactly one of $D_1,D_2$ which means the other is not dense. I am not sure if non-divisible spaces are any easier to come by if we only perfect spaces instead of connected spaces.

Under AC being Hausdorff is necessary. For the cofinite topology is compact and connected and any infinite subset is dense. Without AC $D$ might be an amorphous set.

Answer 1

Here are some details following [1].

Claim: There is a countable dense $X \subseteq [0, 1]^{\mathfrak{c}}$ which does not have a dense co-dense subset.

Proof: Follows from (1) + (2) below.

(1) Every countable dense subspace of $2^{\mathfrak{c}}$ is homeomorphic to a countable dense subspace of $[0, 1]^{\mathfrak{c}}$.

Proof of (1): Use the fact that for every countable dense $D \subseteq 2^{\omega}$, $2^{\omega} \setminus D$ is homeomorphic to the Baire space $([0, 1] \setminus \mathbb{Q})^{\omega}$.

(2) $2^{\mathfrak{c}}$ has a countable dense subspace $X$ which has no dense codense subset.

Proof of (2): (Alas et al. [1]) Let $\{A_i : i < \mathfrak{c}\}$ list all infinite coinfinite subsets of $\omega$. Inductively, try to construct $X_i = \{x_{i, n}: n < \omega\} \subseteq 2^{\mathfrak{c} +i}$ for $i < \mathfrak{c}$ such that the following hold.

(a) $X_i$ is dense in $2^{\mathfrak{c}+i}$.

(b) If $i < j$, then $x_{j, n} \upharpoonright (\mathfrak{c} + i) = x_{i, n}$.

(c) If $\{x_{i, n}: n \in A_i\}$ is dense codense in $X_i$, then $\{x_{i+1, n}: n \in A_i\}$ is open in $X_{i+1}$.

There is no problem at stages $i = 0$, limit.

At stage $i + 1$: If $X_i$ has no dense codense subset, we terminate the construction and put $X = X_i$ - So $X \subseteq 2^{\mathfrak{c} + i} \cong 2^{\mathfrak{c}}$ and we are done. Otherwise, choose an infinite cofinite $A \subseteq \omega$ such that

(i) $\{x_{i, n}: n \in A\}$ is dense codense in $X_i$ and

(ii) if $\{x_{i, n}: n \in A_i\}$ is dense codense in $X_i$, then $A = A_i$

and define $x_{i+1, n} = x_{i, n} \cup \{(i, 1)\}$ if $n \in A$ and $x_{i+1, n} = x_{i, n} \cup \{(i, 0)\}$ if $n \notin A$. As noted above we can assume that the construction can be carried out at every $i < \mathfrak{c}$ and we set $X = \{\bigcup_{i < \mathfrak{c}} x_{i, n} : n < \omega\}$. It is easily checked that $X$ is dense in $2^{\mathfrak{c} + \mathfrak{c}} \cong 2^{\mathfrak{c}}$ and has no dense codense subset.

[1]: Alas et al., Irresolvable and submaximal spaces: Homogeneity versus σ-discreteness and new ZFC examples, Topology and its Applications, Volume 107, Issue 3, 4 November 2000, Pages 259-273

Answer 2

Here is an easier way of getting Ashutosh's item (2), that is a countable dense subset of $2^\mathfrak{c}$ which has no dense co-dense subset.

Let $\mathcal{I}$ be an independent family of subsets of $\omega$. What this means is that for every finite subfamilies $\mathcal{A}$ and $\mathcal{B}$ of $\mathcal{I}$, we have:

$$\bigcap \mathcal{A} \cap \bigcap \{\omega \setminus B: B \in \mathcal{B} \} \neq \emptyset$$.

There is an independent family $\mathcal{I}$ of subsets of $\omega$ having cardinality continuum (see this Mathoverflow post for a dozen proofs of that) and by Zorn's Lemma we can assume that $\mathcal{I}$ is maximal with respect to the property of being independent. Let $\{I_\alpha: \alpha < \mathfrak{c}\}$ be an enumeration of $\mathcal{I}$.

Define a set $D=\{x_n: n \in \mathbb{N}\} \subset 2^\mathfrak{c}$ as follows. For every $\alpha < \mathfrak{c}$

$$x_n(\alpha)=\begin{cases} 1 \textrm{ if } n \in I_\alpha\\ 0 \textrm{ otherwise} \end{cases}$$

Since $\mathcal{I}$ is an independent family, the set $D$ is dense in $2^\mathfrak{c}$ and since $\mathcal{I}$ is a maximal independent family, the space $D$ has no dense co-dense subset. Indeed let, $E \subset D$ be a dense co-dense subset and let $J \subset \omega$ be such that $E=\{x_n: n \in J \}$. Then we must have either $J \notin \mathcal{I}$ or $\omega \setminus J \notin \mathcal{I}$ and adding that set into $\mathcal{I}$ we get a strictly larger independent family, which contradicts maximality of $\mathcal{I}$.