Does there exist a closed, bounded surface $S$ embedded in $\mathbb{R}^3$ that has a geodesic $\gamma$ that spirals around a point $x$, getting closer and closer, but never reaching $x$?

Here I am thinking of the metric on $S$ as inherited from $\mathbb{R}^3$. A portion of such a $\gamma$ might look something like this on the surface, with $\gamma$ confined to a smaller and smaller area on $S$:

Perhaps $S$ is not smooth at $x$.
Answered by Robert Bryant in the comments: A surface of revolution with a "pit" whose bottom is $x$ could support a geodesic that spirals down the pit. "If the geodesic actually limits to $x$, ...[then $x$ cannot be] a smooth point of the surface... However, if the geodesic is instead limiting to a (very small) closed geodesic encircling $x$, ... then $x$ could be a smooth point."

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    $\begingroup$ If you make a small tube spiralling inward, like a hose winding around on your lawn, then you get a geodesic going down the tube, by the reflection symmetry in the horizontal plane (the lawn). You can make the tube get smaller as you go along it. You can put a little half ball on one end of the hose to get a closed set, but of course the limit point is not a smooth point of the surface. $\endgroup$ – Ben McKay Dec 19 '16 at 12:51
  • $\begingroup$ @BenMcKay: Nice example! That spirals in 3D, but is not actually spiraling on the surface, in the sense of being confined to a smaller and smaller surface patch. I tried to rephrase the post to emphasize the latter. $\endgroup$ – Joseph O'Rourke Dec 19 '16 at 13:03
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    $\begingroup$ You need to make 'closer and closer' more precise, by, for example, requiring that the $\omega$-limit of the geodesic be $x$. Otherwise, there are such examples, easily constructed as surfaces of revolution with $x$ as a fixed point. If you do put on this limiting hypothesis, then $S$ cannot be smooth at $x$ (or even $C^1$, I believe, certainly not $C^2$). $\endgroup$ – Robert Bryant Dec 19 '16 at 13:06
  • $\begingroup$ @RobertBryant: May I ask: Is such an example essentially a well or pit which the geodesic spirals down without ever reaching the bottom $x$? $\endgroup$ – Joseph O'Rourke Dec 20 '16 at 14:47
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    $\begingroup$ @JosephO'Rourke: Yes, except that if the geodesic actually limts to $x$, i.e., the bottom of the pit, then $x$ must actually be a very sharp point, in particular, not a smooth point of the surface at all. However, if the geodesic is instead limiting to a (very small) closed geodesic encircling $x$ (which would fit your condition of always coming 'closer and closer', but with a positive lower bound on the distance to $x$), then $x$ could be a smooth point. $\endgroup$ – Robert Bryant Dec 20 '16 at 15:07

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