There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. 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?
 A: 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.
A: Try $\left(\begin{matrix}1\\t\\t^2\end{matrix}\right)$ with $t\in\mathbb R$.
Do you know to compute
$$\left\vert\begin{matrix}1&1&1\\r&s&t\\r^2&s^2&t^2\end{matrix}\right\vert? $$
A: 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 $x,y,z$.
A: 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.)
A: 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.
A: 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.
A: 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.
