A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H =(V, E)$ consists of a set $V$ and a set $E \subseteq {\cal P}(V)$ of subsets of $V$. A *hypergraph coloring* is a map $c: V\to \kappa$, where $\kappa \neq \emptyset$ is a cardinal and the restriction $c\restriction\_e: e \to \kappa$ is non-constant whenever $e$ has more than $1$ element. The *chromatic number* $\chi(H)$ is the least cardinal $\kappa \neq \emptyset$ such that there is a coloring $c: V \to \kappa$.


The map $c:V\to \kappa$ is said to be a *conflict-free* if every (non-empty) edge $e\in E$ contains at least one vertex $v$ of a color unique in $e$, or more formally, if there is $v\in e$ such that $$e \;\cap\; \bigl(c^{-1}(\{c(v)\})\bigr) = \{v\}.$$ The *conflict-free chromatic number* $\chi\_{\text{cf}}(H)$ is the least cardinal $\kappa \neq \emptyset$ such that there is a conflict-free coloring $c: V \to \kappa$.


(Conflict-free colorings were motivated by a frequency assignment problem in cellular networks, see the introduction of [Pach and Tardos - Conflict-free colorings of graphs and hypergraphs](https://www.math.nyu.edu/%7Epach/publications/ConflictFreeGraph052909.pdf).)


Of course, every conflict-free coloring is a coloring in the "traditional" sense, so $\chi(H) \leq \chi\_{\text{cf}}(H)$.


Let $\tau$ be the Euclidean topology on $\mathbb{R}$. The chromatic number $\chi(\mathbb{R}, \tau)$ equals $2$ (see [Chromatic number of a connected Hausdorff space](https://mathoverflow.net/questions/322042/chromatic-number-of-a-connected-hausdorff-space)).


**Question.** What is $\chi\_{\text{cf}}(\mathbb{R},\tau)$?



Conflict-free coloring of $\mathbb{R}$ with the Euclidean topology
By colouring every element of $\Bbb Q$ with a unique colour and every element of $\Bbb R\setminus\Bbb Q$ with the same colour (different from all the colours used so far), we see that $\chi\_\mathrm{cf}(\Bbb R)\leq\aleph\_0$.


But in fact $\aleph\_0$ is a lower bound in any reasonable space.


**Lemma 1:** Let $X$ be a $T\_1$ space with at least three points. Then $\chi\_{\mathrm{cf}}(X)>2$.


**Proof:** Suppose for a contradiction $\chi\_{\mathrm{cf}}(X)=2$ as witnessed by $c\colon X\to 2$. Let $x\in X$ be such that $c(x)\neq c(x')$ for all $x'\in X\setminus\{x\}$. But now we have that $X\setminus\{x\}$ is a monochromatic open set in $X$ with at least two points, a contradiction.


**Lemma 2:** Let $X$ be an infinite $T\_1$ space. Then $\chi\_{\mathrm{cf}}(X)\geq\aleph\_0$.


**Proof:** Suppose for a contradiction $\chi\_{\mathrm{cf}}(X)=n$ as witnessed by $c\colon X\to n$. Let $x\_1\in X$ be such that $c(x)\neq c(x')$ for all $x'\in X\setminus\{x\_1\}$. Now $c\upharpoonright X\setminus\{x\_1\}$ witnesses that $\chi\_{\mathrm{cf}}(X\setminus\{x\_1\})\leq n-1$. Proceed inductively to find $x\_1,\ldots,x\_{n-2}$ so that $c\upharpoonright X\setminus\{x\_1,\ldots,x\_{n-2}\}$ witnesses that the latter space has conflict-free chromatic number at most $2$, which contradicts Lemma 1.


In conclusion $\chi\_{\mathrm{cf}}(\Bbb R)=\aleph\_0$.


