For a given natural number $k$, I'm going to call a subset $T$ of the plane $\Bbb{R}^2$ a $k$-traversal if, for any $x \in \Bbb{R}$, \begin{align*} k &= \operatorname{card} \{(a, b) \in T : a = x\} \\ &= \operatorname{card} \{(a, b) \in T : b = x\} \\ &= \operatorname{card} \{(a, b) \in T : a + b = x\}. \end{align*} It's not difficult to see that lines in the plane that aren't parallel to the $x$-axis, the $y$-axis, or the line $x + y = 0$ will be $1$-traversals (but are far from the only examples!). We can make $2$-traversals without difficulty by taking two disjoint $1$-traversals and unioning them. My question is,
Is there a $2$-traversal that cannot be decomposed into the union of two $1$-traversals?
The idea of (irreducible) $k$-traversals is a concept from latin squares. I'm trying to consider the concept when applied to an (uncountably) infinite Latin square generated by the group $(\Bbb{R}, + )$.
This was a question that popped into my head years ago when attending a combinatorics conference. Combinatorics is not my forte, but it seemed like an interesting question, and I thought I'd share it.
Vincent's example... $T$ is the union of two graphs $y=\frac{|x-1|}{2}$ and $y=-\frac{|x+1|}{2}$.
Each vertical line meets $T$ in two points, one in the top graph and one in the bottom graph.
Each horizontal line meets $T$ in two points: either two in the top graph, or two in the bottom graph, or the two corners on the $x$-axis.
Each line of slope $-1$ meets $T$in two points, one in the top graph and one in the bottom graph.
Thus, $T$ is a $2$-traversal.
Next, we claim it is impossible to have $T = A \cup B$ where $A$ and $B$ are $1$-traversals and $A \cap B = \varnothing$. Indeed, consider the three points $$ u:=\left(0,\frac12\right),\quad v:=\left(2,\frac12\right),\quad w:=\left(2,-\frac32\right). $$
Now $A$ contains exactly one of the points $u,v$, since they are both on the line $y=\frac12$. $A$ contains exactly one of the points $v,w$, since they are both on the line $x=2$. And $A$ contains exactly one of the points $w,u$, since they are both on the line $x+y=\frac12$. Impossible.