= Geodesic image on a compact two-pole surface of revolution is not dense
Let a compact smooth surface of revolution have exactly two poles, and let $c=f(v)^2\dot u=f(v)\cos\theta$ be the <Clairaut first integral for a surface of revolution> of a complete unit-speed <geodesic>. If $c\ne0$, then $f(v)\geq|c|$, so the geodesic avoids open neighbourhoods of both poles. If $c=0$, then $u$ is constant away from the poles and the geodesic lies in a <meridian of a surface of revolution>, whose complement contains a nonempty open set. Thus no complete geodesic has dense image.
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