Let $G=(V,E)$ be a simple, undirected graph such that every vertex has degree at least $2$. Given $n\in\mathbb{N}$, a map $c:E \to \{1,\ldots, n\}$ is said to be a *weak coloring* if for every $v\in V$ the edges adjacent to $v$ do not all have the same color. (More formally, we want the restriction $c|\_{E(v)}$ to be non-constant, where $E(v) = \{e\in E: v\in e\}$.)


[These](https://mathoverflow.net/a/285233/8628) [two](https://mathoverflow.net/a/285234/8628) nice posts by [Mikail Tikhomirov](https://mathoverflow.net/users/106512/mikhail-tikhomirov) and [Brendan McKay](https://mathoverflow.net/users/9025/brendan-mckay) respectively show that for every finite graph there is a weak edge coloring with $3$ colors. I tried to carry through their arguments with transfinite induction to infinite graphs - without success.


**Question.** If $G=(V,E)$ is an infinite simple undirected graph, is there a weak edge coloring $c:E \to \{1,2,3\}$?



Are all infinite graphs $3$-weak-edge colorable?
If $E$ is countable, then we can define the function $c$ by induction as follows. 
Choose an identification $E = \{1,2,\ldots\}$ and choose $c(1) \in \{1,2,3\}$ arbitrarily. Assume that $c(k)$ is defined for every $k < n$. Let $v$ and $v'$ denote the ends of $n$ and let $E(v)$ (resp. $E(v')$) be the edges containing $v$ (resp. $v'$). There are at most two forbidden colors for $n$:
if edges in $E(v)$ which have already been colored have the same colour $i\_v$, then $i\_v$ is possibly a forbidden color. Same for $v'$ and we have another possibly forbidden color $i\_{v'}$. Choose $c(n) \in \{1,2,3\} \backslash \{i\_v, i\_{v'}\}$.


