The famous [De Bruijn–Erdős theorem](https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory)) and its [hypergraphs generalization](https://mathoverflow.net/questions/432595/de-bruijn-erd%C5%91s-theorem-for-hypergraphs) states the following.


**Theorem.** *Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **finite** and that for some $k \in \mathbb{N}$, every **finite** sub-hypergraph of $H=(V,E)$ can be properly colored with $k$ colors. Then $H$ can also be properly colored with $k$ colors.*


It looks like that the standard proof based on [Tychonoff's theorem](https://en.wikipedia.org/wiki/Tychonoff%27s_theorem) doesn't work when the edges are allowed to be infinite. So, I wonder if the following analogue of De Bruijn–Erdős theorem holds.



> 
> Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **countable**. Moreover, assume that for some $k \in \mathbb{N}$ and for every **countable** $W \subset V$, the hypergraph $H(W)= (W,E(W))$ can be properly colored with $k$ colors, where $E(W) = \{e \in E: e \subset W\}$. Then $H=(V,E)$ can also be properly colored with $k$ colors.
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Erdős and Hajnal proved in [[EH]](https://link.springer.com/article/10.1007/BF02066676) the following related result on families of bounded intersections.


**Theorem.** *Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **countably infinite**. Moreover, assume that there exists $m \in \mathbb{N}$ such that $|e\_1\cap e\_2|<m$ for all $e\_1,e\_2 \in E$. Then $H=(V,E)$ can be properly colored with $2$ colors.*


Can this result be strengthened as follows to deal with all families of finite intersections?



> 
> Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **countably infinite**, and that $e\_1\cap e\_2$ is finite for all $e\_1,e\_2 \in E$. Then $H=(V,E)$ can be properly colored with $2$ colors.
> 
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[EH] P. Erdős & A. Hajnal, *On a property of families of sets*, Acta Math. Academiae Scientiarum Hungarica, 12, 87–123 (1964).



Two questions on infinite hypergraphs
The answer to the first question is also no, by a minor modification of the proof of the Elekes-Hoffmann result cited in the answer to the second question. In fact, we get the following:


**Theorem** There is a set $V$ of cardinality $2^{\aleph\_0}$ and a collection $E$ of countably infinite subsets of $V$ such that, for every countable $W \subset V$, $H(W) = (W, E(W))$ is 2-colorable, but $H=(V,E)$ cannot be properly colored with countably many colors.


**Proof** Let $V$ be the collection of all functions of the form $f:\beta \rightarrow \omega$, where $\beta$ is a countable ordinal. For each such $f$, let $A\_f$ be the set of all $n < \omega$ such that the preimage $f^{-1}(\{n\})$ is infinite. For each $n \in A\_f$, let $e\_{f,n} := \{f\} \cup \{f \restriction \alpha \mid \alpha \in f^{-1}(\{n\})\}$. Then let $E\_f := \{e\_{f,n} \mid n \in A\_f\}$. Note that each element of $E\_f$ is countably infinite and has a maximal element, namely $f$ itself. Finally, let $E := \bigcup\_{f \in V} E\_f$.


Let us first show that $H(W)$ is 2-colorable for every countable $W \subset V$. The point is that, for each such $W$, $E(W)$ is countable, since, by construction, $E(W) \subseteq \bigcup\_{f \in W} E\_f$, and each $E\_f$ is countable. Therefore, one can enumerate both $W$ and $E(W)$ in order-type $\omega$ and use these enumerations to inductively define a coloring $c:W \rightarrow \{0,1\}$, diagonalizing against the elements of $E(W)$ to make sure none of them are monochromatic for $c$ (here we use the fact that every element of $E(W)$ is infinite).


We finally show that $H$ cannot be properly colored with countably many colors. Fix a function $c:V \rightarrow \omega$. We will show that it is not a proper coloring of $H$. Let us define a function $g:\omega\_1 \rightarrow \omega$ by recursively specifying $g \restriction \beta$ for $\beta \leq \omega\_1$. If $\beta \leq \omega\_1$ is a limit ordinal and we have specified $g \restriction \alpha$ for all $\alpha < \beta$, then $g \restriction \beta = \bigcup\_{\alpha < \beta} g \restriction \alpha$. For the successor step, suppose that we have constructed $g \restriction \beta$. To define $g \restriction (\beta + 1)$, we must define $g(\beta)$; do so by setting $g(\beta) = c(g \restriction \beta)$.


Let $A$ be the set of all $n < \omega$ such that $g^{-1}(\{n\})$ is unbounded in $\omega\_1$, and let $\beta < \omega\_1$ be large enough so that:


* for all $\alpha < \omega\_1$, if $g(\alpha) \notin A$, then $\alpha < \beta$; and
* for all $n \in A$, $g^{-1}(\{n\}) \cap \beta$ is infinite.


Let $n = g(\beta)$, and let $f = g \restriction \beta$. By construction, we have $n = c(g \restriction \beta) = c(f)$. By the first bullet point above, we know that $n \in A$. By the second bullet point, we know that $f^{-1}(\{n\})$ is infinite, so $n \in A\_f$ and $e\_{f,n} = \{f\} \cup \{f \restriction \alpha \mid \alpha \in f^{-1}(\{n\})$ is in $E$. But for every $\alpha \in f^{-1}(\{n\})$, we have $c(f \restriction \alpha) = c(g \restriction \alpha) = g(\alpha) = n = c(f)$, so $e\_{f,n}$ is monochromatic for $c$, showing that $c$ is not a proper coloring of $H$.


