# Can a countable dense subset be split into two disjoint dense subsets?

Suppose $X$ is a compact connected Hausdorff space and $D \subset X$ countable and dense. Can we always write $D=D_1 \cup D_2$ as a disjoint union of countable dense subsets? More generally if $U \subset X$ is open and $D \subset U$ is countable and dense does there exist a decomposition $D=D_1 \cup D_2$ with each $D_1 \cap U$ and $D_2 \cap U$ dense in $U$?

The result is easy to show if $X$ has a countable basis $U_1,U_2,\ldots$. We just use how each $U_n \cap D$ is infinite to assign distinct elements of each $U_n$ to each of $D_1$ and $D_2$ and proceed by induction.

$X$ being connected, or at least perfect, cannot be dropped. For if $x \in D$ is isolated any dense set includes it, and $x$ must be an element of exactly one of $D_1,D_2$. So I wouldn't be surprised if the result fails for $U$ but holds for $X$.

• If $D$ is dense in $X$ and $U$ is open in $X$ then $\overline U=\overline {U\cap D}$ so $U\cap D$ is dense in $U$. So if $D=D_1\cup D_2$ where $D_1$ and $D_2$ are dense in $X$ then $U\cap D_1$ and $U\cap D_2$ are dense in $U.$ – DanielWainfleet Apr 18 '18 at 14:48
• If it is true then it holds for any separable connected locally-compact Tychonoff space $U$ because if $U$ is not already compact then $U$ has a 1-point compactification $X=U\cup \{p\}.$ I still don't have an answer. – DanielWainfleet Apr 18 '18 at 14:52
• This has been asked answered and accepted at mathoverflow.net/questions/298069/… – Dap Apr 23 '18 at 18:18

Below is Ashutosh's answer from MathOverflow:

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

• I think you should accept this answer to move this question off the "unanswered" list. – Noah Schweber Feb 26 '19 at 0:58