Infinite set of non-collinear points with property

Question

Let $S \subset \mathbb{R}^{2}$ be an infinite collection of non-collinear points with the following properties:
1.) For any $a \in \mathbb{R}$, $|S \cap \{ (x,a) : x \in \mathbb{R} \}| = |S \cap \{ (a,x) : x \in \mathbb{R}\}| = 1$.
2.) For $(x_{0}, y_{0})$, $(x_{1} , y_{1}) \in S$ such that $x_{0} < x_{1}$, we have that $y_{0} < y_{1}$.
Must there exist an infinite subset $T \subset S$ such that $\forall x \in T$, $x$ is not a convex combination of points in $T \setminus \{ x\}$?

Answer 1

I’m assuming that the non-collinearity assumption is actually that no three points of $S$ are collinear. Define $c:S^{(3)}\to[2]$ as follows: if $\{p_0,p_1,p_2\}\in S^{(3)}$, where $p_k=\langle x_k,y_k\rangle$ for $k=0,1,2$ and $x_0<x_1<x_2$, then

$$c(\{p_0,p_1,p_2\})=\begin{cases} 0,&\text{if }\frac{y_1-y_0}{x_1-x_0}<\frac{y_2-y_1}{x_2-x_1}\\ 1,&\text{otherwise}\;. \end{cases}$$

Let $T$ be an infinite subset of $S$ such that $c$ is constant of $T^{(3)}$. Suppose that $c\left[T^{(3)}\right]=\{0\}$; the other case is similar. For $p=\langle x_0,y_0\rangle\in T$ let

$$m(p)=\sup\left\{\frac{y_0-y}{x_0-x}:\langle x,y\rangle\in T\text{ and }x<x_0\right\}\;;$$

then $T\setminus\{p\}$ lies in the closed half-plane lying above the line through $p$ of slope $m(p)$ and has at most one point on that line, and it follows that $p$ is not a convex combination of points of $T\setminus\{p\}$.

Answer 2

$S$ is obviously the graph of a monotone bijective function $f:\mathbb{R}\rightarrow\mathbb{R}$. This implies that $f$ is continuous, see here. A continuous function that does not have three collinear points on its graph must be either strictly convex or strictly concave (see excursion below). This means that none of the points of $S$ is a convex combination of any other points of $S,$ which in turn means that each subset $T\subset S$ has the property "$\forall x\in T,$ $x$ is not a convex combination of points in $T\setminus\{x\}.$" It does not even have to be an infinite subset or a particular subset.

Excursion: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function that is neither strictly convex on its entire domain nor strictly concave on its entire domain. Then the graph $S$ of $f$ contains three collinear points.

I will call three points $(x_1,y_1),\;\ldots,\;(x_3,y_3)$ a convex triple if $x_1 < x_2 < x_3$ and $(x_3-x_1)\,y_2 < (x_3-x_2)\,y_1 + (x_2-x_1)\,y_3,$ which means that $(x_2,y_2)$ is located below the line through $(x_1,y_1)$ and $(x_3,y_3).$

Likewise, the points $(x_1,y_1),\;\ldots,\;(x_3,y_3)$ are called a concave triple if $x_1 < x_2 < x_3$ and $(x_3-x_1)\,y_2 > (x_3-x_2)\,y_1 + (x_2-x_1)\,y_3$ which means that $(x_2,y_2)$ is located above the line through $(x_1,y_1)$ and $(x_3,y_3).$

Four points $P_1,\ldots,P_4$ form a convex (concave) quadruple if $P_1,$ $P_2,$ $P_3$ as well as $P_2,$ $P_3,$ $P_4$ form convex (concave) triples.

Four points $P_1,\ldots,P_4$ form a zigzag quadruple if $P_1,$ $P_2,$ $P_3$ form a convex triple and $P_2,$ $P_3,$ $P_4$ form a concave triple or the other way round.

Suppose $f$ is a function as described above. In addition, suppose there were no three collinear points on the graph $S$ of $f.$ (This assumption avoids the necessity to deal with collinearity each time we talk about three or more points in $S.$)

Our goal now is to find a zigzag quadrupel on $S.$ As $f$ is neither strictly convex nor strictly concave, and $S$ does not contain collinear points, we can find a convex triple $(x_1,f(x_1)),\ldots,(x_3,f(x_3))$ as well as a concave triple $(w_1,f(w_1)),\ldots,(w_3,f(w_3)).$

If $w_3 > x_1,$ we choose $x_4 > \max(x_3,w_3).$ If $(x_1,f(x_1)),\;\ldots,\;(x_4,f(x_4))$ is a zigzag quadruple, we have found what we are looking for. Otherwise, $(x_1,f(x_1)),\ldots,(x_4,f(x_4))$ must be a convex quadruple, and we can make $x_4$ our new $x_3,$ $x_3$ our new $x_2$ and $x_2$ our new $x_1.$ We repeat this until we either found a zigzag quadruple or until we achieved $x_1 > w_3.$

If we have not found a zigzag quadrupel yet, we now have at least a convex triple and a concave triple such the the convex one is located completely more to the right than the concave one. In this case, we look at the four points $(w_2,f(w_2)),\;(w_3,f(w_3)),\;(x_1,f(x_1)),\;(x_2,f(x_2)).$ If they are a zigzag quadruple, we have found what we are looking for. If they are a convex quadruple, then $(w_1,f(w_1)),\;(w_2,f(w_2)),\;(w_3,f(w_3)),\;(x_1,f(x_1))$ is obviously a zigzag quadruple. If they are a concave quadruple, then $(w_3,f(w_3)),\;(x_1,f(x_1)),\;(x_2,f(x_2)),\;(x_3,f(x_3))$ is obviously a zigzag quadruple.

In either case, we now have proven the existence of a zigzag quadruple $(a_1,b_1),\ldots,\;(a_4,b_4)$ on $S.$

Now we show that there is a line such that the points of the zigzag quadruple are above the line and below the line in alternating order. Start with a line through $(a_2,b_2)$ and $(a_3,b_3).$ $(a_1,b_1)$ will be on one side of the line and $(a_4,b_4)$ on the other side. Now rotate the line a little bit around $\left(\frac{a_2+a_3}{2},\frac{b_2+b_3}{2}\right)$ such that the line approaches $(a_1,b_1)$ as well as $(a_4,b_4)$ without hitting one of them. Now the first point and the third point are on one side of the line, while the second point and the fourth point are on the other side.

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be the function that defines this line. $f-g$ is a continuous function. Apply the intermediate value theorem to $f-g$ for each of the three intervals $[a_1,a_2],$ $[a_2,a_3]$ and $[a_3,a_4]$ and you get three crossings between $f$ and $g.$ This shows that there are three collinear points on the graph of $f,$ which completes the proof.