Here is a question I asked myself years ago. Since it is not really in my field, I hope to find some (partial) answers here... Since it was unclear, I precise that I am looking for an answer in ZFC, so using the axiom of choice if needed.

The chromatic number of a graph $G$ is the minimal cardinal of a partition of $G$ into independant sets, i.e. sets all of whose connected components are trivial. This definition of independant sets suggests that a good analogue of them in the setting of topological spaces would be totally disconected sets. This motivate the following definition:

Definition.Define the chromatic number of a topological space $X$, denoted by $\chi(X)$, as the minimal cardinal of a partition of $X$ into totally disconnected sets.

My main question is the following:

Question.What is the chromatic number of $\mathbb{R}^n$?

Some remarks. If $X$ can be embedded in $Y$, then $\chi(X) \leqslant \chi(Y)$. We also have that $\chi(X \times Y) \leqslant \chi(X)\chi(Y)$, because if $(A_i)_{i \in I}$ and $(B_j)_{j \in J}$ are respective partitions of $X$ and $Y$ into totally disconnected sets, then $(A_i \times B_j)_{(i, j) \in I \times J}$ is so for $X \times Y$. We also have that $\chi(\mathbb{R}) = 2$, witnessed by the partition $\{\mathbb{Q}, \mathbb{I}\}$, where $\mathbb{I}$ denote the set of irrational numbers. So $\chi(\mathbb{R}^n) \leqslant 2^n$.

But we have better. Actually, $\chi(\mathbb{R}^2) \leqslant 3$, witnessed by the partition $\{\mathbb{Q}^2, (\mathbb{Q} \times \mathbb{I} \cup \mathbb{I}\times \mathbb{Q}), \mathbb{I}^2\}$ (where $\mathbb{Q} \times \mathbb{I} \cup \mathbb{I}\times \mathbb{Q}$ is totally disconnected because it embeds into $\mathbb{I}^2$ via $(x, y) \mapsto (x + y, x - y)$). So $\chi(\mathbb{R}^{2n}) \leqslant 3^n$ and $\chi(\mathbb{R}^{2n + 1}) \leqslant 2 \times 3^n$.

My conjecture is that $\chi(\mathbb{R}^n) = n + 1$. For the upper bound, I suspect that some partition of the form $\{A_0, \ldots A_n\}$ where $A_i$ is the set of elements of $\mathbb{R}^n$ having exactly $i$ rational coordinates could work, but I don't manage to prove anything. For the lower bound, I only have an intuition given by the following "image": if you consider $A \subseteq \mathbb{R}^n$ totally disconnected, then it sounds reasonable to think that there exists a set $B\subseteq \mathbb{R}^n$, homeomorphic to $\mathbb{R}^{n - 1}$, passing "between" the points of $A$; however, this is just an inutuition and I don't know if it is true. I suspect that some algebraic topology would be needed to prove that, but I have almost no experience in this field so I couldn't investigate more; I am not even able to show that $\chi(\mathbb{R}^n) > 2$ for some $n$.

An other possibility would be that $\chi(\mathbb{R}^n) = 2$ for every $n$, because of some "weird" colouring. What makes me suspect that is that if we replace "connected" by "arcwise connected" in the definition of the chromatic number, then it becomes easy to build a partition of $\mathbb{R}^n$ into two parts each of which having no non-trivial arcwise connected subset, by a diagonal argument using the axiom of choice. However, this proof uses the fact that there are exactly $\mathfrak{c}$ arcs $\gamma : [0, 1] \longrightarrow \mathbb{R}^n$. To do the same proof for connected sets, we would need the existence of a family of $\mathfrak{c}$ non-trivial connected subsets of $\mathbb{R}^n$ such that every non-trivial connected subset of $\mathbb{R}^n$ has a subset in this family, and I don't think that such a family exists (but I don't know).

In case where we manage to prove that $\chi(\mathbb{R}^n) = 2$ thanks to some construction using the axiom of choice, it would be interesting to investigate what happens (still supposing choice) if we impose some restriction on the complexity of the sets in the partition, for example to be Borel.

withoutchoice? $\endgroup$ – Asaf Karagila♦ Feb 5 '18 at 14:245more comments