Crossings in an Eulerian Trail Exercise 11.2 in Graph Theory by Harary says

Every plane eulerian graph contains an eulerian trial that never crosses itself.

What does it mean for a trail to not cross itself?  The book does not give a formal definition of this notion.
 A: I don't know the "formal" definition, but informally it means just what you would think. If you regard the Eulerian trail as a curve in the plane, the curve does not cross itself, in the sense that the graphs of $y=0$ and $y=x$ cross at the origin, but the graphs of $y=0$ and $y=x^2$ touch without crossing.
For instance, consider the plane Eulerian graph with vertices $v=(0,0)$, $w=(1,0)$, $x=(1,1)$, $y=(-1,-1)$, $z=(-1,0)$, and straight-line edges $vw,wx,xv,vy,yz,zv$. The Eulerian trail $z,v,w,x,v,y,z$ crosses itself at $v$, but the Eulerian trail $z,v,x,w,v,y,z$ does not cross itself.
P.S. Following a suggestion by the OP, here's an attempt at defining a self-crossing for an Eulerian trail in a plane graph $G$. The trail crosses itself at a vertex $v$ if, among the edges of $G$ that are incident with $v$, there are four distinct edges $a,b,c,d$ which occur in the cyclic order $(a,b,c,d)$ in the clockwise ordering of the edges incident with $v$, and such that the edges $a$ and $c$ are traversed consecutively (though not necessarily in that order) in the trail, and the same goes for the edges $b$ and $d$.
A: A planar trail "crosses itself" if you cannot draw it without edges crossing.  Connecting at a vertex does not count as crossing.  
