Can every large point set be connected to a given knot? Let $K$ be a given knot, and
$P$ a set of points in $\mathbb{R}^3$ in general position,
general position in the sense that no three points are collinear
and no four coplanar.
Define the point-set knot number of $K$, $z(K)=n$,
as the smallest number $n$ such that
every point set $P$ of at least $n$ points, $|P| \ge n$,
may be connected to a
non-self-intersecting polygonal knot equivalent to $K$.
Say that $z(K)=\infty$ if there is no such $n$.
The vertices of the polygon must be exactly $P$:
every point in $P$ is used,
and the polygon does not turn at a point not in $P$.
For example, $z(K_0)=3$ for $K_0$ the unknot.
Here is why. Let the points in $P$ be $p_1, p_2, \ldots, p_n$ sorted top
to bottom vertically. Form the polygon $(p_1, p_2, \ldots, p_n, p_1)$,
closing the chain by connecting $p_n$ to $p_1$.
It is clear that the only possible self-intersection that can occur
involves the last segment $s=p_n p_1$. Suppose $s$ passes through a vertex.
Then three points are collinear.
Suppose $s$ passes through an edge (e.g., $(4,5)$ below). Then four points are coplanar.
Both cases violate the general position assumption.
So the polygon is non-self-intersecting.
And it should be clear it is equivalent
to the unknot.

                   


Obviously $z(K)$ is at least the stick number of $K$.
But the real question is whether $z(K)$ is finite.
 A: The problem has been answered (positively) by Negami [1]: His first main observation was that for any sufficiently large $n$, by a Ramsey Theory argument, every set of $n$ points in general position in $\mathbb{R}^3$ contains $n$ points on an order $n$ curve (without loss of generality: the moment curve); the second part of the argument shows that every knot occurs on a sufficiently large set of points on the moment curve. So $z(K)$ is finite for all knots $K$.
The Ramsey theory argument needed for this is also presented in Björner et al. [2, Prop. 9.4.7].
An answer for the trefoil knot was given by Ramirez Alfonsin [3]: Any set of $7$ points in general position contains a trefoil or its mirror image. He denotes this as $m(T)=7$.
It seems to me that more than $7$ points are needed if one insists to get the trefoil, and not the mirror image.
[1] Seiya Negami: "Ramsey theorems for knots, links and spacial graphs", Trans. Amer. Math. Soc. (2)324 (1991), 527-541 http://www.ams.org/journals/tran/1991-324-02/S0002-9947-1991-1069741-9/S0002-9947-1991-1069741-9.pdf
[2] Anders Björner, Michel Las Vergnas, Bernd Sturmfels, Neil White & Günter M. Ziegler: "Oriented Matroids", Cambridge University Press 1993.
[3] Jorge L. Alfonsin: "Spacial graphs and oriented matroids: the trefoil", Discrete Comput. Geometry 22 (1999), 149-158. http://link.springer.com/article/10.1007/PL00009446
A: Here is a plan, based a result I proved in my undergraduate (!) thesis!
Let $Y = (\mathbb{R}^2\times \{ 0\}) \cup (\{0 \} \times \mathbb{R} \times [0,\infty))$.
I'll embed the knot $K$ in $Y$, like so:


*

*First, take your knot $K$ draw it mostly in the plane, with nice tidy semicircular hoops for the overcrossings.

*Then isotope the plane so that all the crossings are in line along the
$y$-axis and the hoops are coplanar poke up off the line.


Now we have a copy of $K$ embedded in $Y$.  The nice thing about this is that we can encode this knot as a list of points with three families of "joining curves".  Thus we can draw an $n$-crossing knot $K$ using $2n$ points to anchor the hoops, $n$ points between the anchor pairs and $3n$ points to act as turnaround points for the "joining curves"  (a total of $6n$ points).
So now I think we can replicate this using general position points.


*

*We may assume that no two points have the same $z$-coordinate.  

*Choose a cylinder with axis parallel to the $z$-axis so that there are $2n$ points inside the cylinder.  

*Divide space into three pie wedges containing $3n$  (I think ${3\over2} n$ would suffice) points each and connect away!


EDIT:  There is a difficulty about how to choose the final connections carefully enough. I'll think about this and leave it up in case it gives anyone else an idea. 
