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> Given a finite connected graph, let $A$ be a set of edges such that each edge in $A$ is not part of a cycle. Suppose that no path contains all edges in $A$. Must it be true that for some three edges in $A$, no path contains all the three edges?
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This is equivalent to showing that if every subset of three edges in $A$ is contained in a path, then there is a path containing all edges in $A$. I’ve tried doing induction, but even going from three to four doesn’t seem obvious. Each set of edges $\{e\_1,e\_2,e\_3\}$ and $\{e\_2,e\_3,e\_4\}$ has a path containing it, but other edges maybe used in both of these paths in various orders.



Three edges in a path
Extend $A$ to a spanning tree $T$ of $G$, which is possible greedily for any acyclic subgraph of a connected graph. Since $A$ cannot be covered by a single path, there is a vertex $v$ such that at least $3$ of the branches of $T$ at $v$ lead to components $C\_1, C\_2, C\_3$ containing edges $a\_1, a\_2, a\_3$ of $A$. We claim that these edges cannot be covered by a single path.


Indeed, suppose there were such a path $P$. Fix an orientation of $P$ and label each $a\_i$ according to whether $P$ traverses $a\_i$ towards or away from $v$. By reversing the orientation and relabelling if necessary we may assume that $P$ traverses $a\_1$ and $a\_2$ away from $v$. But then there must be a path from $C\_1$ to the start point of $a\_2$ 
not passing through $a\_1$, which would place $a\_1$ on a cycle in $G$, a contradiction.


