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Could anyone just give me hint for this one?

There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. True or false?

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Do you know any facts about linearly independent sets? –  Chris Eagle Dec 12 '12 at 22:27
Is there a subset of $\mathbb{R}^3$ with the property that every plane through the origin meets it in at most 2 points? –  Hurkyl Nov 28 '13 at 13:54
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7 Answers

up vote 9 down vote accepted

Three vectors in $\mathbb{R}^3$ are linearly dependent if and only if they lie in a plane.

Consider the following process for building $S$. We can start with the empty set, and choose any two vectors $v_1, v_2 \in \mathbb{R}^3$ and add them to $S$. Then to choose a third vector $v_3$ to add to $S$, we must make sure it is not in the unique plane containing (i.e. spanned by) $v_1$ and $v_2$. Thus $v_3$ can be any vector in $\mathbb{R}^3 \backslash span(v_1, v_2)$.

Similarly, if at some stage $S = \{v_1, \ldots, v_k\}$, we can add to $S$ any vector $v_{k+1}$ in $\mathbb{R}^3 \backslash \bigcup_{x_i, x_j} span(x_i, x_j)$. Note that $\bigcup_{x_i, x_j} span(x_i, x_j)$ is a finite union of planes, so it can never be all of $\mathbb{R}^3$. In this way we can choose an infinite set with the desired property.

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The key issue with this approach is to prove that a finite union of planes is not all of $\mathbb R^3$. For a more general result: mathoverflow.net/questions/26/… –  Andres Caicedo Dec 12 '12 at 22:47
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Try $\left(\begin{matrix}1\\t\\t^2\end{matrix}\right)$ with $t\in\mathbb R$. Do you know to compute $$\left\vert\begin{matrix}1&1&1\\r&s&t\\r^2&s^2&t^2\end{matrix}\right\vert? $$

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good one!!. Please clear my doubt, consider the random gaussian vector $[x,y,z]$, where each entries are independent gaussian random variables. So shouldn't the set of all realizations of this random vector be a set of the kind he is looking for? –  dineshdileep Dec 13 '12 at 5:44
the determinant is $(st^2-s^2t)-(rt^2-tr^2)+(rs^2-sr^2)$ –  Bunuelian Trick Dec 14 '12 at 5:08
@Kuttus ... and that equals $(r-s)(s-t)(t-r)$ and is non-zero unless two of the numbers are equal. :) Actually, the fact that the functions $x\mapsto 1$, $x\mapsto x$ and $x\mapsto x^2$ are linearly independent lurks behind this as a shortcut: If a linear combinaton of the rows is the zero vector, then the correspondnig quadratic polynomial has three distinct roots $r,s,t$, hence is the zero polynomal, hence the linear combination was in fact the trivial combination, hence the rows are linearly independent, hence so are the columns (which is what we were actually after). –  Hagen von Eitzen Dec 14 '12 at 11:39
+1 I'd just like to point out to future readers of this excellent answer that the curve $t\to (1,t,t^2)$ in $\mathbb{R}^3$ is also known as the twisted cubic. –  Amitesh Datta Jul 27 '13 at 5:42
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Consider vectors of the form $v_x=(1,x,x^2)^T$. Then for any distinct $x,y,z\in \mathbb R$ matrix $(v_x v_yv_z)$ is nonsingular (Vandermonde matrix), so $v_x$, $v_y$, $v_z$ are linearly independent.

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The parametric curve $t\mapsto(1,t,t^2)$ has the property that any three distinct points on it are linearly independent (without the "distinct" one clearly cannot have a solution). To check, just compute the determinant, which is Vandermonde.

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Inspired by the Vandermonde matrix example, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly convex function. Then $$\left\vert\begin{matrix}1&1&1\\x&y&z\\f(x)&f(y)&f(z)\end{matrix}\right\vert\not=0$$ for any three distinct numbers $x,y,z$.

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The elegant example given by Adam can be generalized: let $P \subset \mathbb R^3$ be any plane that does not pass through the origin (equipped with the usual subspace topology), and let $C \subset P$ be a strictly convex* closed subset of $P$. Then the boundary $S = \partial_P\, C$ of $C$ in $P$ satisfies your requirement.

(* By "strictly convex", I mean that any non-trivial convex combination of two points in $C$ should belong to the interior of $C$, i.e. for all $x,y \in C$ and all $0 < \alpha < 1$, the point $z = \alpha x + (1-\alpha)y$ lies in the interior of $C$. Geometrically, this basically means that no part of the boundary should be a straight line segment.)

Proof: Let $x$ and $y$ be any two distinct points in $S$, and let $z$ linear combination of $x$ and $y$ which is distinct from both. By definition, $z = \alpha x + \beta y$ for some $\alpha$ and $\beta$. If $\alpha + \beta \ne 1$, then $z \notin P$, and thus $z \notin S$. Thus, $z$ can only belong to $P$ if $\beta = 1 - \alpha$. Let us assume below that this is the case.

If $0 < \alpha < 1$, by strict convexity as defined above, $z \in \operatorname{int} C$ and thus $z \notin S$. If $\alpha \in \{0,1\}$, then $z = x$ or $z = y$, contradicting the assumption of distinctness. The only remaining possibility is that $\alpha < 0$ or $\alpha > 1$, but if either of those holds, then either $x$ is a convex combination of $z$ and $y$, or $y$ is a convex combination of $z$ and $x$. In either case, $z$ cannot belong to $S \subset C$, since otherwise the fact that $x,y \in S = \partial\,C = C \setminus \operatorname{int} C$ would contradict the assumption that $C$ is strictly convex.

Besides the example given by Adam, one example possible concrete example of such a set would be e.g. the circle given by $P = \{(x,y,z) \in \mathbb R^3: z = 1\}$, $C = \{(x,y,z) \in P: x^2 + y^2 \le 1\}$ and $S = \partial_P\,C = \{(x,y,z) \in P: x^2 + y^2 = 1\}$.

The sets given by constructions like mine and Adam's are generally one-dimensional curves. Indeed, it's not hard to show that a set satisfying your requirements cannot contain any two-dimensional surface, since any surface would have to intersect some plane passing through the origin along a curve containing more than two points.

However, it is possible to construct a set which satisfies your requirements and is dense in $\mathbb R^3$. Basically, to do this, you'd start with an arbitrary enumeration $\langle a_i \rangle_{i \in \mathbb N}$ of the rational points $\mathbb Q^3$ (which are a countable dense subset of $\mathbb R^3$) and then, for each $i \in \mathbb N$, choose a point $b_i \in \mathbb R^3 \setminus \{0\}$ such that $|a_i - b_i| < 1/i$ and such that $b_i$ is pairwise linearly independent of all $b_j$, $j < i$. (This is always possible, since the set of pairwise linear combinations of $b_j$, $j < i$ is the intersection of finitely many planes, and thus has zero volume.) Then let $S = \{b_i: i \in \mathbb N\}$.

Note that the "construction" outlined above is not entirely constructive, since it requires an arbitrary choice of a point at each step of an infinite process. I do, however, believe that this dependency on choice could be eliminated by giving an explicit rule that assigns a definite value to $b_i$, given $a_i$ and $\{b_j: j < i\}$, similarly to the method I sketched in this answer to a related question.

(Do note the caveat brought up by Asaf Karagila in the comments to the other answer, though. I still think the construction I outlined shouldn't require even dependent choice, since an explicit formula for $b_i$ may be provided: basically, the idea is that, for each $i$, it should be possible to explicitly generate a finite ordered list of points $c_k$, $k \in \{1, \dotsc, n\}$ within radius $1/i$ of $a_i$, such that at least one of those points is guaranteed to be pairwise linearly independent of $\{b_j: j < i\}$, and then always choose the first such point. Still, I freely admit that I'm not nearly as experienced with this stuff as Asaf is, and that there could be some gap that I've missed.)

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Existence : Consider three points $x_1,\ x_2,\ x_3$ in a plane $P\ :\ x+y+z=1$ which are not in a line.

Pick $x_4$ in $P-\bigcup L_{ij}$ where $L_{ij}$ is a line passing $x_i,\ x_j\ (1\leq i,\ j \leq 3)$.

Repeat this pocess infinitely.

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