One common definition of a ruled surface $S$ is that, through every point $p \in S$, there passes a line $L(p)$ that lies in $S$: $L(p) \subset S$. My question is:

Q0. Is there any loosening of "through every point $p$" that leads to the conclusion that $S$ is nevertheless still ruled?

(Added:) Let us assume $S$ is smooth and $L(p)$ is an infinite line (not a segment). More specifically:

Q1. Is there a surface $S$ through which all but a finite set of exceptional points, passes a line that lies in $S$? So it is not ruled, but there are only finite exceptions?

Q2. Is there a surface $S$ through which all but a countably infinite set of exceptional points, passes a line that lies in $S$?

Q3. If a set of points $P$ is dense on $S$, and through every $p\in P$ there passes a line that lies in $S$, does that imply that $S$ ruled?

  Image from Univ Arizona Math Teaching Tools
(Added 14Oct14). Here is Robert Bryant's example surface $S$ that shows the answer to Q1 is—remarkably—Yes. From $z=xy$ he removes the four red points $$(1,0,0),\; (−1,0,0),\; (0,−1,0),\; (0,1,0)\;,$$ and then there is no complete line through the five yellow points: $$(0,0,0),\; (1,1,1),\; (1,−1,−1),\; (−1,1,−1),\; (−1,−1,1)\;,$$ but there is a line through every other point of $S$.

  • $\begingroup$ What level of regularity are you willing to assume about $S$? For example, do you assume that $S$ is smooth? Also, are you interested in pieces of surfaces, i.e., in 'locally ruled' surfaces that may contain only line segments rather than whole lines? $\endgroup$ – Robert Bryant Oct 11 '14 at 10:33
  • $\begingroup$ @RobertBryant: I was primarily interested in smooth $S$ and containing infinite lines. I will so edit. Thanks. $\endgroup$ – Joseph O'Rourke Oct 11 '14 at 11:55
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    $\begingroup$ It seems to me that if a sequence of lines in $S$ converges to a line $L$, then (assuming $S$ is topologically closed) $L$ must also be in $S$. Doesn't this answer all your questions? $\endgroup$ – Benoît Kloeckner Oct 11 '14 at 18:00
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    $\begingroup$ @Joseph: However, you did not state the assumption that $S$ be topologically closed in your original question, and I assumed that you did not want to. Without this assumption, the answer to Q1 is yes: Take $S$ to be the hyperboloid $z-xy=0$ minus the four points $(1,0,0)$, $(-1,0,0)$, $(0,-1,0)$, and $(0,1,0)$. Then, through every point of $S$, there passes a (complete) line in $S$ except for the $5$ points $(0,0,0)$, $(1,1,0)$, $(1,-1,0)$, $(-1,1,0)$, and $(-1,-1,0)$. $\endgroup$ – Robert Bryant Oct 14 '14 at 10:54
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    $\begingroup$ @JosephO'Rourke: By the way, I probably should have mentioned that if, instead, one takes $S$ to be the hyperboloid $z-xy=0$ minus the points $(1,1,1)$ and $(-1,-1,1)$, then except for the two points $(-1,1,-1)$ and $(1,-1,-1)$, every point of $S$ lies on a (complete) line in $S$. I don't know whether one can construct an example with only one exceptional point. $\endgroup$ – Robert Bryant Oct 14 '14 at 11:43

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