Proof. Let $V\subset [n]^d$ such that $|V| = n^{d - \delta }$ . Set $r = \frac {k}{2} + 1$ and $E_r = \binom {V}{r}$ to be the collection of $r$ -tuples of $V$ . Notice that the sum of an $r$ -tuple from $V$ belongs to $[rn]^d$ . For each $v \in [rn]^d$ , we define

\begin{equation*} E_r(v)=\{\{v_1,\ldots ,v_r\}\in E_r\,:\, v_1+\ldots +v_r=v\}. \end{equation*}

Then for $T_1,T_2 \in E_r(v)$ , where $T_1 = \{v_1,\ldots , v_{r}\}$ and $T_2 = \{u_1,\ldots , u_{r}\}$ , we have

\begin{equation*} v_1 + \cdots + v_{r} = v = u_1 + \cdots + u_{r}, \end{equation*}

which implies that $T_1\cup T_2$ lies on a common $k$ -flat. Let

\begin{equation*} E_{2r} = \bigcup _{v \in [rn]^d}\ \bigcup _{T_1,T_2 \in E_r(v)} \{T_1, T_2\}. \end{equation*}

Hence, for each $\{T_1, T_2\} \in E_{2r}$ , $T_1\cup T_2$ lies on a $k$ -flat. Moreover, by Jensen’s inequality, we have

\begin{equation*} |E_{2r}| = \sum _{v \in [rn]^d} \binom {|E_r(v)|}{2} \geq (rn)^d \binom { \frac {\sum _{v } |E_r(v)| }{ (rn)^d}}{2} = (rn)^d \binom { |E_r|/ (rn)^d}{2} \geq \frac {|E_r|^2}{4(rn)^d}. \end{equation*}

Since $k$ and $d$ are fixed and $r = \frac {k}{2} + 1$ and $|V|= n^{d - \delta }$ ,

\begin{equation*} |E_r|^2 = \binom {|V|}{r}^2 = \binom {|V|}{(k/2) + 1}^2 \geq \Omega (n^{(k + 2)(d-\delta )}). \end{equation*}

Combining the two inequalities above gives

\begin{equation*} |E_{2r}| \geq \Omega (n^{(k + 1)d - (k + 2)\delta }). \end{equation*}

We say that $\{T_1, T_2\} \in E_{2r}$ is good if $T_1\cap T_2 = \emptyset$ , and the $(k + 2)$ -tuple $(T_1\cup T_2)$ is non-degenerate. Otherwise, we say that $\{T_1,T_2\}$ is bad. In what follows, we will show that at least half of the pairs (i.e. elements) in $E_{2r}$ are good. To this end, we will need the following claim.

Proof of Claim. Write $T_1 = \{v_1,\ldots , v_{r}\}$ and $T_2 = \{u_1,\ldots , u_{r}\}$ . Let us consider the following cases.

Case 1. Suppose $T_1\cap T_2 \neq \emptyset$ . Then, without loss of generality, there is an integer $j \lt r$ such that

\begin{equation*} v_1 + \cdots + v_j = u_1 + \cdots + u_j, \end{equation*}

where $v_1,\ldots ,v_j,u_1,\ldots ,u_j$ are all distinct elements, and $v_t = u_t$ for $t\gt j$ . Thus, $|T_1\cup T_2| = 2j + (r-j)$ . The $2j$ elements above lie on a $(2j - 2)$ -flat. Adding the remaining $r-j$ points implies that $T_1\cup T_2$ lies on a $(j-2 + r)$ -flat. Since $r = \frac {k}{2} + 1$ and $j \leq \frac {k}{2},$ $T_1\cup T_2$ lies on a $(k-1)$ -flat.

Case 2. Suppose $T_1\cap T_2 = \emptyset$ . Then $T_1\cup T_2$ must be degenerate, which means there is a subset $S\subset T_1\cup T_2$ of $j$ elements such that $S$ lies on a $(j-2)$ -flat, for some $3 \leq j \leq k + 1$ . Without loss of generality, we can assume that $v_1 \not \in S$ . Hence, $(T_1\cup T_2)\setminus \{v_1\}$ lies on a $(k-1)$ -flat. On the other hand, we have

\begin{equation*} v_1 = u_1+\cdots + u_{r} -v_2 -\cdots - v_{r}. \end{equation*}

Hence, $v_1$ is in the affine hull of $(T_1\cup T_2)\setminus \{v_1\}$ which implies that $T_1\cup T_2$ lies on a $(k-1)$ -flat.

Proof of Claim. For the sake of contradiction, suppose at least half of the pairs in $E_{2r}$ are bad. Let $H$ be the collection of all the $j$ -flats spanned by subsets of $V$ for all $j\leq k-1$ . Notice that if $S\subset V$ spans a $j$ -flat $h$ , then $h$ is also spanned by only $j+1$ elements from $S$ . So we have

\begin{equation*} |H| \leq \sum _{j=0}^{k-1}|V|^{j+1} \leq k n^{k(d - \delta )}. \end{equation*}

For each bad pair $\{T_1, T_2\} \in E_{2r}$ , $T_1\cup T_2$ lies on a $j$ -flat from $H$ by Claim 2.2. By the pigeonhole principle, there is a $j$ -flat $h$ with $j\leq k-1$ such that at least

\begin{equation*} \frac {|E_{2r}|/2}{|H|} \geq \frac {\Omega (n^{(k + 1)d - (k + 2)\delta })}{2kn^{k(d - \delta )}} = \Omega (n^{d - 2\delta }) \end{equation*}

bad pairs from $E_{2r}$ have the property that their union lies in $h$ . On the other hand, since $h$ contains at most $n^{k-1}$ points from $[n]^d$ , $h$ can correspond to at most $O(n^{(k-1)(k + 2)})$ bad pairs from $E_{2r}$ . Since we assumed $d - 2\delta \gt (k-1)(k + 2)$ , we have a contradiction for $n$ sufficiently large.

Proof. If $U$ spans a $j$ -flat for some $j\lt i-1$ , then by definition no non-degenerate $(k+2)$ -tuple contains $U$ . Hence we can assume $U$ spans a $(i-1)$ -flat. Observe that a non-degenerate $(k+2)$ -tuple $T$ , which lies on a $k$ -flat and contains $U$ , must contain a $(k+1)$ -tuple $T'\subset T$ such that $T'$ spans a $k$ -flat and $U\subset T'$ . Then there are at most $n^{(k + 1 - i)(d-\delta )}$ ways to add $k + 1 - i$ points to $U$ from $V$ to obtain such $T'$ . After $T'$ is determined, there are at most $n^k$ ways to add a final point from the affine hull of $T'$ to obtain $T$ . So we conclude the proof by multiplication.

Proof. We count the number of ways to choose an $(\ell +s)$ -tuple $T$ that spans a $j$ -flat. There are at most $n^{(j+1)d}$ ways to choose a subset $T'\subset T$ of size $j+1$ that spans the affine hull of $T$ . After this $T'$ is determined, there are at most $n^{(\ell +s-1-j)j}$ ways to add the remaining $\ell +s-1-j$ points from the $j$ -flat spanned by $T'$ . Then the total number of $(\ell +s)$ -tuples that lie on an $\ell$ -flat is at most

\begin{equation*} \sum _{j=1}^{\ell } n^{(j+1)d+(\ell +s-1-j)j}\leq \sum _{j=1}^{\ell } n^{(j+1)d+(\ell +s-1-j)\ell } \leq \sum _{j=1}^{\ell } n^{(\ell +1)d+(s-1)\ell } \leq \ell \cdot n^{(\ell +1)d+(s-1)\ell }, \end{equation*}

where the second inequality uses $\ell \leq d$ .

Proof of Theorem1.1 . In dimensions $d' \geq 3$ where $d'$ is odd, we apply Theorem3.2 with $k = \ell = d' - 1$ to obtain a point set $V$ of size $N$ in $\mathbb{R}^d$ with the property that no $d' +2$ members lie on a $(d' - 1)$ -flat, and every subset of size $\Omega \left ( N^{\frac {1}{2} + \frac {1}{2d'} + \epsilon } \right )$ contains $d' + 1$ members on a $(d' -1)$ -flat. By projecting $V$ to a generic $d'$ -dimensional subspace of $\mathbb{R}^d$ , we obtain $N$ points in $\mathbb{R}^{d'}$ with no $d' + 2$ members on a common hyperplane, and every subset in general position has size $O\left ( N^{\frac {1}{2} + \frac {1}{2d'} + \epsilon } \right )$ .

In dimensions $d' \geq 4$ where $d'$ is even, we apply Theorem3.2 with $k = d'- 2$ and $\ell = d' -1$ to obtain a point set $V$ of size $N$ in $\mathbb{R}^d$ with the property that no $d' +2$ members on a $(d'-1)$ -flat, and every subset of size $\Omega \left ( N^{\frac {1}{2} + \frac {1}{d' - 1} + \epsilon } \right )$ contains $d'$ members on a $(d' -2)$ -flat. By adding another point from this subset, we obtain $d' + 1$ members on a $(d' - 1)$ -flat. Hence, by projecting to $V$ a generic $d'$ -dimensional subspace of $\mathbb{R}^d$ , we obtain $N$ points in $\mathbb{R}^{d'}$ with no $d' + 2$ members on a common hyperplane, and every subset in general position has size $O\left ( N^{\frac {1}{2} + \frac {1}{d' - 1} + \epsilon } \right )$ .

Since $\epsilon$ is arbitrary and $N$ grows to infinity, we can conclude the proof of Theorem1.1 after renaming $d'$ to $d$ .

Proof of Theorem3.2 . Let $d$ be a sufficiently large integer and $n$ tend to infinity. We denote $\mathcal{H}$ as the hypergraph with $V(\mathcal{H})=[n]^d$ and $E(\mathcal{H})$ consisting of non-degenerate $(k+2)$ -tuples $T$ such that $T$ lies on a $k$ -flat. We shall construct a rooted tree $\mathfrak{T}$ whose nodes are labelled with vertex subsets of $\mathcal{H}$ as follows. We start with $\mathfrak{T}$ consisting of one root node labelled with $V(\mathcal{H})$ . Iteratively, if there is a leaf $x \in \mathfrak{T}$ whose labelled set $C_x$ has size at least $n^{\frac {k}{k + 1}d + k}$ , we apply Lemma 3.1 to $\mathcal{H}[C_x]$ with $\tau = n^{-\frac {k}{k + 1}d + \delta + \epsilon }$ where $\delta$ is defined by $|C_x| = n^{d - \delta }$ . As a consequence, Lemma 3.1 produces a collection $\mathcal{C}$ of subsets of $C_x$ . Then we create a child of $x$ in $\mathfrak{T}$ labelled by $C$ for each $C \in \mathcal{C}$ . The iteration continues until there is no leaf $x\in \mathfrak{T}$ with $|C_x| \geq n^{\frac {k}{k + 1}d + k}$ .

During the interative construction of $\mathfrak{T}$ , we need to verify the hypothesis of Lemma 3.1, that is,

\begin{equation*} \Delta _i(\mathcal{H}[C_x]) \leq c \cdot \tau ^{i-1} \frac {|E(\mathcal{H}[C_x])|}{|V(\mathcal{H}[C_x])|} \quad \text{for all }2\leq i\leq k+2. \end{equation*}

To check this, we use Lemma 2.4 to upper bound $\Delta _i(\mathcal{H}[C_x])$ for $2\leq i \lt k+2$ and use the trivial bound $\Delta _i(\mathcal{H}[C_x])\leq 1$ for $i = k+2$ . On the other hand, we use Lemma 2.1 to lower bound $|E(\mathcal{H}[C_x])|$ . We shall use $n^{d - \delta } = |V(\mathcal{H}')| \geq n^{\frac {k}{k + 1}d + k}$ as well. Since this is a straightforward computation, whose detail will be given as Claim 4.2 in the proof of Theorem1.2, we skip it here.

Now, we analyse this rooted tree $\mathfrak{T}$ . According to Lemma 3.1(c), if $y$ (labelled with $C_y$ ) is a child of $x$ (labelled with $C_x$ ) in $\mathfrak{T}$ , the number of edges induced by $C_y$ shrinks from that by $C_x$ by a constant factor $(1-c)$ . On the other hand, a reasonably large set induces many edges in $\mathcal{H}$ by Lemma 2.1 (assuming $d$ is large). This means the height of $\mathfrak{T}$ is upper bounded by $O(\log n)$ , and in particular our iterative construction ends. According to Lemma 3.1(b), the number of children of any node $x$ in $\mathfrak{T}$ is at most

\begin{equation*} |\mathcal{C}| \leq \exp ({c^{-1} \cdot \tau |C_x| \cdot \log (1/\tau )}) \leq \exp \left (O\left (n^{\frac {d}{k + 1} + \epsilon }\cdot \log {n}\right )\right ). \end{equation*}

Therefore, let $\mathfrak{C}$ be the collection of sets labelling the leaves of $\mathfrak{T}$ . Hence, we have

\begin{equation*} |\mathfrak{C}| \leq \exp \left (O\left (n^{\frac {d}{k + 1} + \epsilon }\cdot \log ^2{n}\right )\right ) \quad \text{and}\quad |C| \leq n^{\frac {k}{k + 1}d + k} \ \text{for all }C\in \mathfrak{C}. \end{equation*}

Furthermore, if $I$ is an independent set of $\mathcal{H}$ that is contained in a vertex subset $C_x$ labelling a non-leaf node $x$ , then by the construction of $\mathfrak{T}$ and Lemma 3.1(a), there exists a child $y$ of $x$ in $\mathfrak{T}$ whose labelling set $C_y$ contains $I$ . This implies every independent set of $\mathcal{H}$ is contained in some member of $\mathfrak{C}$ . Elements in this collection $\mathfrak{C}$ are called containers.

Next, we randomly select a subset of $[n]^d$ by keeping each point independently with probability $p$ . Let $S$ be the set of selected elements. Then for each $(\ell + 3)$ -tuple $T$ in $S$ that lies on an $\ell$ -flat, we delete one point from $T$ . We denote the resulting set of points by $S'$ . By Lemma 2.5, we have

\begin{equation*} \mathbb{E}[|S'|] \geq pn^d - p^{\ell +3}\ell n^{(\ell +1)d+2\ell }. \end{equation*}

By setting $p=(2\ell )^{-\frac {1}{\ell +2}}n^{-\frac {\ell }{\ell +2}(d+2)}$ , we have

\begin{equation*} \mathbb{E}[|S'|] \geq \frac {pn^d}{2} =\Omega \left (n^{\frac {2(d-\ell )}{\ell +2}}\right ). \end{equation*}

Finally, we set $m = n^{\frac {d}{k + 1} + 2\epsilon }$ . Let $X$ denote the number of independent sets of $\mathcal{H}$ in $S'$ with cardinality $m$ . Using the family of containers, we have

\begin{align*} \mathbb{E}[X] & \leq |\mathfrak{C}|\cdot \binom {n^{\frac {k}{k + 1}d + k}}{m} \cdot p^{m} \\ &\leq \exp \left (O\left (n^{\frac {d}{k + 1} + \epsilon }\cdot \log ^2{n}\right )\right ) \cdot \left (\frac {e \cdot n^{\frac {k}{k + 1}d + k}}{m}\right )^m p^m \\ &\leq \exp \left (O\left (n^{\frac {d}{k + 1} + \epsilon }\cdot \log ^2{n}\right )\right ) \cdot \left (e \cdot n^{\frac {k-1}{k + 1}d + k -2\epsilon }\right )^m \left ((2\ell )^{-\frac {1}{\ell +2}} \cdot n^{-\frac {\ell }{\ell +2}(d+2)}\right )^m \\ &\leq \exp \left (O\left (n^{\frac {d}{k + 1} + \epsilon }\cdot \log ^2{n}\right )\right ) \cdot \left (\frac {1}{2}\right )^m\\ &\leq o(1). \end{align*}

Here, the fourth inequality uses the following consequence of $k\leq \ell$ and $d$ being large:

\begin{equation*} \frac {k-1}{k + 1}d + k -2\epsilon -\frac {\ell }{\ell +2}(d+2) \lt 0. \end{equation*}

Notice that $|S'|$ is exponentially concentrated around its mean by Chernoff’s inequality. Therefore, some realisation of $S'$ satisfies: $|S'|=N=\Omega (n^{2(d-\ell )/(\ell + 2)})$ ; $S'$ contains no $(\ell +3)$ -tuples on a $\ell$ -flat; and $\mathcal{H}[S']$ does not contain an independent set of $\mathcal{H}$ with cardinality

\begin{equation*} m = n^{\frac {d}{k + 1} + 2\epsilon } = O\left ( N^{\frac {\ell + 2}{2(k + 1)} + \frac {(\ell +2)\ell }{2(k+1)(d-\ell )} + \frac {\ell + 2}{d-\ell }\epsilon }\right ) \leq O\left ( N^{\frac {\ell + 2}{2(k + 1)} + \epsilon }\right ). \end{equation*}

Here, we assume $d = d(\epsilon ,k,\ell )$ is sufficiently large so that

\begin{equation*} \frac {(\ell +2)\ell }{2(k+1)(d-\ell )} + \frac {\ell + 2}{d-\ell }\epsilon \leq \epsilon . \end{equation*}

Notice that $S'$ not containing an independent set of size $m$ means every subset of $S'$ of size $m$ contains $k+2$ points on a $k$ -flat. We conclude the proof by renaming $S'$ to $V$ .

Proof. Just as before, let $\mathcal{H}$ be the hypergraph with $V(\mathcal{H})=[n]^d$ and $E(\mathcal{H})$ consisting of non-degenerate $(k+2)$ -tuples $T$ such that $T$ lies on a $k$ -flat. We let $q=q(k,r,s)$ be a quantity that will be determined later. We again construct a rooted tree $\mathfrak{T}$ whose nodes are labelled with vertex subsets of $\mathcal{H}$ . We start with $\mathfrak{T}$ consisting of one root node labelled with $V(\mathcal{H})$ . Iteratively, if there is a leaf $x \in \mathfrak{T}$ whose labelled set $C_x$ has size at least $n^{qd + k}$ , we apply Lemma 3.1 to $\mathcal{H}' = \mathcal{H}[C_x]$ with $\tau = n^{-qd + \delta + \epsilon }$ where $\delta$ is defined by $|C_x| = n^{d - \delta }$ . We shall use the claim below to verify the hypothesis of Lemma 3.1. As a consequence, Lemma 3.1 produces a collection $\mathcal{C}$ of subsets of $C_x$ . Then we create a child of $x$ in $\mathfrak{T}$ labelled by $C$ for each $C \in \mathcal{C}$ . The iteration continues until there is no leaf $x\in \mathfrak{T}$ with $|C_x| \geq n^{qd + k}$ .

Claim 4.2. If $\frac {1}{2} \lt q \leq \frac {k}{k+1}$ and $\mathcal{H}'$ defined as above, then

\begin{equation*} \Delta _i(\mathcal{H}') \leq c \cdot \tau ^{i-1} \frac {|E(\mathcal{H}')|}{|V(\mathcal{H}')|} \quad \text{for all }2\leq i\leq k+2, \end{equation*}

where $c$ is the constant in Lemma 3.1 depending only on $k$ .

Proof of Claim. First, we notice that

(4.1) \begin{equation} n^{d - \delta } = |V(\mathcal{H}')| \geq n^{qd + k} \implies \delta \leq d - qd - k. \end{equation}

Assuming $d$ is large, we have $|E(\mathcal{H}')| \geq \Omega (n^{(k + 1)d - (k + 2)\delta })$ by Lemma 2.1.

For $2\leq i \lt k+2$ , Lemma 2.4 gives us $\Delta _i(\mathcal{H}') \leq n^{(k+1-i)(d-\delta )+k}$ . Hence, it suffices to check

\begin{equation*} n^{(k+1-i)(d-\delta ) + k} \ll \left (n^{-qd+\delta +\epsilon }\right )^{i-1} \cdot \frac {n^{(k+1)d-(k+2)\delta }}{n^{d-\delta }}. \end{equation*}

Simplifying and comparing the exponents over $n$ , this is implied by

\begin{equation*} (i-1) d + k + (i-1)\epsilon \gt (i-1) q d + \delta . \end{equation*}

Since $d$ is sufficiently large, it suffices to compare the coefficients of $d$ . Applying (4.1) and simplifying the terms, the inequality above is implied by $i-1 \geq (i-2)q + 1$ , which is true by our hypothesis.

For $i = k+2$ , we have $\Delta _i(\mathcal{H}') \leq 1$ trivially. Hence, it suffices to check

\begin{equation*} 1 \ll \left (n^{-qd+\delta +\epsilon }\right )^{k+1} \cdot \frac {n^{(k+1)d-(k+2)\delta }}{n^{d-\delta }}. \end{equation*}

Simplifying and comparing the exponents over $n$ , this is implied by

\begin{equation*} (k+1)qd \lt kd + (k+1)\epsilon . \end{equation*}

Again, since $d$ is sufficiently large, it suffices to compare the coefficients of $d$ . The inequality above is implied by $(k+1) q \leq k$ , which is true by our hypothesis.

We can analyse this rooted tree $\mathfrak{T}$ using arguments similar to the previous section. We can conclude that there exists a collection $\mathfrak{C}$ of vertex subsets of $\mathcal{H}$ with

\begin{equation*} |\mathfrak{C}| \leq \exp \left (O\left (n^{d-qd + \epsilon }\cdot \log ^2{n}\right )\right ) \quad \text{and}\quad |C| \leq n^{qd + k} \ \text{for all }C\in \mathfrak{C}. \end{equation*}

and every independent set of $\mathcal{H}$ is contained in some member of $\mathfrak{C}$ .

Next, we randomly select a subset of $[n]^d$ by keeping each point independently with probability $p$ . Let $S$ be the set of selected elements. Then for each $(\ell + s)$ -tuple $T$ in $S$ that lies on an $\ell$ -flat, we delete one point from $T$ . We denote the resulting set of points by $S'$ . By Lemma 2.5, we have

\begin{equation*} \mathbb{E}[|S'|] \geq pn^d - p^{\ell +s}\ell n^{(\ell +1)d+(s-1)\ell }. \end{equation*}

By setting $p=(2\ell )^{-\frac {1}{\ell +s-1}}n^{-\frac {\ell }{\ell +s-1}(d+s-1)}$ , we have

\begin{equation*} \mathbb{E}[|S'|] \geq \frac {pn^d}{2} =\Omega \left (n^{\frac {(s-1)(d-\ell )}{\ell +s-1}}\right ). \end{equation*}

Finally, we set $m = n^{d-qd + 2\epsilon }$ . Let $X$ denote the number of independent sets of $\mathcal{H}$ in $S'$ with cardinality $m$ . With a foresight soon to be self-evident, we choose

(4.2) \begin{equation} q = \frac {1}{2} \cdot \frac {2\ell +s-1}{\ell +s-1} + \frac {1}{2d} \cdot \frac {\ell (s-1)}{\ell +s-1} -\frac {k}{2d}. \end{equation}

We remark that our hypothesis on $k,\ell ,s$ implies $\frac {1}{2}\lt q \leq \frac {k}{k + 1}$ assuming $d$ is large, hence Claim 4.2 can be applied in construction of $\mathfrak{T}$ .

Using the family $\mathfrak{C}$ , we can estimate

\begin{align*} \mathbb{E}[X] & \leq |\mathfrak{C}|\cdot \binom {n^{qd + k}}{m} \cdot p^{m} \\ &\leq \exp \left (O\left (n^{d-qd + \epsilon }\cdot \log ^2{n}\right )\right ) \cdot \left (\frac {e \cdot n^{qd + k}}{m}\right )^m p^m \\ &\leq \exp \left (O\left (n^{d-qd + \epsilon }\cdot \log ^2{n}\right )\right ) \cdot \left (e \cdot n^{(2q-1)d + k -2\epsilon }\right )^m \left ((2\ell )^{-\frac {1}{\ell +s-1}}n^{-\frac {\ell }{\ell +s-1}(d+s-1)}\right )^m \\ &\leq \exp \left (O\left (n^{d-qd + \epsilon }\cdot \log ^2{n}\right )\right ) \cdot \left (\frac {1}{2}\right )^m\\ &\leq o(1). \end{align*}

Here, the fourth inequality uses the following consequence of (4.2):

\begin{equation*} (2q-1)d + k -2\epsilon -\frac {\ell }{\ell +s-1}(d+s-1) \lt 0. \end{equation*}

Notice that $|S'|$ is exponentially concentrated around its mean by Chernoff’s inequality. Therefore, some realisation of $S'$ satisfies: $|S'|=N=\Omega \left (n^{\frac {(s-1)(d-\ell )}{\ell +s-1}}\right )$ ; $S'$ contains no $(\ell +s)$ -tuples on a $\ell$ -flat; and $\mathcal{H}[S']$ does not contain an independent set of $\mathcal{H}$ with cardinality

\begin{equation*} m = n^{d - qd + 2\epsilon } = O\left ( N^{\frac {1}{2}+ \left (\frac {k}{2} + 2\epsilon \right )\cdot \frac {\ell +s-1}{(s-1)(d-\ell )}} \right ) \leq O\left ( N^{\frac {1}{2} + \epsilon }\right ). \end{equation*}

Here, we assume $d = d(\epsilon ,k,\ell ,s)$ is sufficiently large so that

\begin{equation*} \left (\frac {k}{2} + 2\epsilon \right )\cdot \frac {\ell +s-1}{(s-1)(d-\ell )} \leq \epsilon . \end{equation*}

Since $S'$ does not contain an independent set of size $m$ , every subset of $S'$ of size $m$ contains $k+2$ points on a $k$ -flat. We conclude the proof by renaming $S'$ to $V$ .

Proof of Theorem 1.2 . In dimensions $d'\geq 3$ where $d'$ is odd, we obtain an upper bound for $\alpha _{d',s'}(N)$ with $d's' + 2 \gt 2d' + 2s'$ . We set $k = \ell = d'-1$ and $s = s' + 1$ , so we can verify $\frac {2\ell +s-1}{\ell +s-1} \lt \frac {2k}{k+1}$ . Hence we can apply Theorem4.1 to obtain a point set $V$ of size $N$ in $\mathbb{R}^{d}$ with the property that no $d' +s'$ members lie on a $(d' - 1)$ -flat, and every subset of size $\Omega (N^{\frac {1}{2} + \epsilon })$ contains $d' + 1$ members on a $(d' -1)$ -flat. By projecting $V$ to a generic $d'$ -dimensional subspace of $\mathbb{R}^d$ , we obtain $N$ points in $\mathbb{R}^{d'}$ with no $d' + s'$ members on a common hyperplane, and every subset in general position has size $O( N^{\frac {1}{2} + \epsilon })$ .

In dimensions $d' \geq 4$ where $d'$ is even, we obtain an upper bound for $\alpha _{d',s'}(N)$ with $d's' + 2 \gt 2d' + 3s'$ . We set $k = d'- 2$ , $\ell = d' -1$ , and $s = s' + 1$ , so we can verify $\frac {2\ell +s-1}{\ell +s-1} \lt \frac {2k}{k+1}$ . Hence we can apply Theorem4.1 to obtain a point set $V$ of size $N$ in $\mathbb{R}^d$ with the property that no $d' +s'$ members on a $(d'-1)$ -flat, and every subset of size $\Omega (N^{\frac {1}{2} + \epsilon })$ contains $d'$ members on a $(d' -2)$ -flat. By adding another point from this subset, we obtain $d' + 1$ members on a $(d' - 1)$ -flat. Hence, by projecting to $V$ a generic $d'$ -dimensional subspace of $\mathbb{R}^d$ , we obtain $N$ points in $\mathbb{R}^{d'}$ with no $d' + s'$ members on a common hyperplane, and every subset in general position has size $O( N^{\frac {1}{2} + \epsilon })$ .

Since $\epsilon$ is arbitrary and $N$ grows to infinity, we can conclude the proof of Theorem1.2 after renaming $d'$ to $d$ and $s'$ to $s$ .

Proof of Theorem5.1 . Let $d$ , $r$ , and $V$ be as given in the hypothesis. Let $m \geq 1$ be an integer that will be determined later. We define

\begin{equation*} S_r = \{v_1 + \cdots +v_{r}\,:\, v_i \in V, v_i \neq v_j\}, \end{equation*}

and a function

\begin{equation*} \sigma :\,\binom {V}{r}\rightarrow S_r,\ \{v_1,\ldots , v_r\} \mapsto v_1 + \cdots + v_r. \end{equation*}

Notice that $\sigma$ is a bijection. Indeed, suppose on the contrary that

\begin{equation*} v_1 + \cdots + v_{r} = v'_1 + \cdots + v'_{r} \end{equation*}

for two different $r$ -tuples in $V$ . Then by setting $(\textbf {x}_1,\ldots ,\textbf {x}_r)=(v_1,\ldots ,v_r)$ , $(\textbf {x}_{r+1},\ldots ,\textbf {x}_{2r})=(v'_1,\ldots ,v'_r)$ , $(\textbf {x}_{2r+1},\ldots ,\textbf {x}_{3r})=(\textbf {x}_{3r+1},\ldots ,\textbf {x}_{4r})$ arbitrarily, and $c_1=c_2=1$ , we obtain a non-trivial solution to (5.2), which is a contradiction. In particular, we have $|S_r| = \binom {|V|}{r}$ .

For $j \in [m]$ and $w \in \mathbb{Z}_j^d$ , we let

\begin{equation*} U_{j,w} = \{u \in \mathbb{Z}^d\,:\, ju + w \in S_r\}. \end{equation*}

Notice that for fixed $j \in [m]$ , we have

\begin{equation*} \sum _{w \in \mathbb{Z}_j^d} |U_{j,w}| = \sum _{w \in \mathbb{Z}_j^d} |\{v \in S_r \,:\, v \equiv w \text{ mod }j\}| = |S_r|. \end{equation*}

Applying Jensen’s inequality to above, we have

(5.3) \begin{equation} \sum _{w \in \mathbb{Z}_j^d} |U_{j,w}|^2 \geq |S_r|^2/j^d. \end{equation}

For $i \geq 0$ , we define

\begin{equation*} \Phi ^i_{U_{j,w}-U_{j,w}}(x) = \{(u_1,u_2)\in \Phi _{U_{j,w}-U_{j,w}}(x)\,:\, |\sigma ^{-1}(ju_1+w)\cap \sigma ^{-1}(ju_2+w)| = i\}. \end{equation*}

It’s obvious that these sets form a partition of $\Phi _{U_{j,w}-U_{j,w}}(x)$ . We also make the following claims.

Claim 5.3. For a fixed $x\in \mathbb{Z}^d$ , we have

\begin{equation*} \sum _{j \in [m]}\sum _{w \in \mathbb{Z}_j^d } |\Phi ^0_{U_{j,w}-U_{j,w}}(x)| \leq 1, \end{equation*}

Proof. For the sake of contradiction, suppose the summation above is at least two, then we have $(u_1,u_2)\in \Phi ^0_{U_{j,w}-U_{j,w}}(x)$ and $(u_3,u_4)\in \Phi ^0_{U_{j',w'}-U_{j',w'}}(x)$ such that either $(u_1,u_2)\neq (u_3,u_4)$ or $(j,w)\neq (j',w')$ .

Let $s_1,s_2,s_3,s_4 \in S_r$ such that $s_1 = ju_1 + w$ , $s_2 = ju_2 + w$ , $s_3 = j'u_3 + w'$ , $s_4 = j'u_4 + w'$ and write $\sigma ^{-1}(s_i)=\{v_{i,1},\ldots ,v_{i,r}\}$ . Notice that $u_1 - u_2 = x = u_3 - u_4$ . Putting these equations together gives us

(5.4) \begin{equation} j'((v_{1,1} + \cdots + v_{1,r}) - (v_{2,1} + \cdots + v_{2,r})) = j((v_{3,1} + \cdots + v_{3,r}) - (v_{4,1} + \cdots + v_{4,r})). \end{equation}

It suffices to show that (5.4) can be seem as a non-trivial solution to (5.2). The proof now falls into the following cases.

Case 1. Suppose $j \neq j'$ . Without loss of generality we can assume $j'\gt j$ . Notice that $(u_1,u_2)\in \Phi ^0_{U_{j,w}-U_{j,w}}(x)$ implies

\begin{equation*} \{v_{1,1},\ldots ,v_{1,r}\}\cap \{v_{2,1},\ldots ,v_{2,r}\}=\emptyset . \end{equation*}

Then after combining like terms in (5.4), the coefficient of $v_1^1$ is at least $j'-j$ , which means this is indeed a non-trivial solution to (5.2).

Case 2. Suppose $j = j'$ , then we must have $s_1 \neq s_3$ . Indeed, if $s_1=s_3$ , we must have $w=w'$ (as $s_1$ modulo $j$ equals $s_3$ modulo $j'$ ) and $s_2=s_4$ (as $j'(s_1-s_2)=j(s_3-s_4)$ ). This is a contradiction to either $(u_1,u_2)\neq (u_3,u_4)$ or $(j,w)\neq (j',w')$ .

Given $s_1 \neq s_3$ , we can assume, without loss of generality, $v_{1,1}\not \in \{v_{3,1},\ldots ,v_{3,r}\}$ . Again, we have $\{v_{1,1},\ldots ,v_{1,r}\}\cap \{v_{2,1},\ldots ,v_{2,r}\}=\emptyset$ . Hence, after combining like terms in (5.4), the coefficient of $v_1^1$ is positive and we have a non-trivial solution to (5.2).

Claim 5.4. For a finite set $T \subset \mathbb{Z}^d$ , and fixed integers $i,j\geq 1$ , we have

\begin{equation*} \sum _{w\in \mathbb{Z}_j^d}\sum _{x\in \mathbb{Z}^d} |\Phi ^{i}_{U_{j,w}-U_{j,w}}(x)|\cdot |\Phi _{T-T}(x)|\leq |V|^{2r-i}|T|. \end{equation*}

Proof. The summation on the left-hand side counts all (ordered) quadruples $(u_1,u_2,t_1,t_2)$ such that $(u_1,u_2)\in \Phi ^{i}_{U_{j,w}-U_{j,w}}(t_1-t_2)$ . For each such a quadruple, let $s_1,s_2 \in S_r$ such that

\begin{equation*} s_1 = ju_1 + w \quad \text{ and}\quad s_2 = ju_2 + w. \end{equation*}

There are at most $|V|^{2r-i}$ ways to choose a pair $(s_1,s_2)$ satisfying $|\sigma ^{-1}(s_1)\cap \sigma ^{-1}(s_2)|=i$ . Such a pair $(s_1,s_2)$ determines $(u_1,u_2)$ uniquely. Moreover, $(s_1,s_2)$ also determines the quantity

\begin{equation*} t_1-t_2=u_1-u_2=\frac {s_1-w}{j}-\frac {s_2-w}{j}=\frac {1}{j}(s_1-s_2). \end{equation*}

After such a pair $(s_1,s_2)$ is chosen, there are at most $|T|$ ways to choose $t_1$ and this will also determine $t_2$ . So we conclude the claim by multiplication.

Now, we set $T = \mathbb{Z}_\ell ^d$ for some integer $\ell$ to be determined later. Notice that $U_{j,w} + T \subset \{0,1,\ldots , \lfloor rn/j\rfloor + \ell -1\}^d$ , which implies

(5.5) \begin{equation} |U_{j,w} + T|\leq (rn/j+\ell )^d. \end{equation}

By Lemma 5.2, we have

\begin{equation*} \frac {|U_{j,w}|^2||T|^2}{|U_{j,w} + T|} \leq \sum _{x \in \mathbb{Z}^d}|\Phi _{U_{j,w}-U_{j,w}}(x)|\cdot |\Phi _{T-T}(x)|. \end{equation*}

Summing over all $j \in [m]$ and $w \in \mathbb{Z}_j^d$ , and using Claims 5.3 and 5.4, we can compute

\begin{align*} \sum _{j\in [m]}\sum _{w \in \mathbb{Z}_j^d}\frac {|U_{j,w}|^2||T|^2}{|U_{j,w} + T|} & \leq \sum _{j\in [m]}\sum _{w \in \mathbb{Z}_j^d} \sum _{x \in \mathbb{Z}^d}|\Phi _{U_{j,w}-U_{j,w}}(x)|\cdot |\Phi _{T-T}(x)| \\ & = \sum _{x \in \mathbb{Z}^d} \sum _{j\in [m]}\sum _{w\in \mathbb{Z}_j^d}\left ( |\Phi ^0_{U_{j,w}-U_{j,w}}(x)| + \sum _{i = 1}^{r}|\Phi ^i_{U_{j,w}-U_{j,w}}(x)|\right )|\Phi _{T-T}(x)| \\ & \leq \sum _{x \in \mathbb{Z}^d}|\Phi _{T-T}(x)| \sum _{j\in [m]}\sum _{w\in \mathbb{Z}_j^d}|\Phi ^0_{U_{j,w}-U_{j,w}}(x)| + \sum _{j\in [m]} \sum _{i = 1}^{r}|V|^{2r-i}\ell ^d \\ & \leq \sum _{x \in \mathbb{Z}^d}\Phi _{T-T}(x) + \sum _{j\in [m]} \sum _{i = 1}^{r-1}|V|^{2r-i}\ell ^d\\ & \leq \ell ^{2d} + rm|V|^{2r-1}\ell ^d, \end{align*}

On the other hand, using (5.3) and (5.5), we can compute

\begin{align*} \sum _{j\in [m]}\sum _{w \in \mathbb{Z}_j^d}\frac {|U_{j,w}|^2||T|^2}{|U_{j,w} + T|} & \geq \sum _{j\in [m]}\sum _{w \in \mathbb{Z}_j^d}\frac {|U_{j,w}|^2\ell ^{2d}}{(rn/j + \ell )^d} \\ & \geq \sum _{j\in [m]} \frac {|S_r|^2\ell ^{2d}}{j^d(rn/j + \ell )^d} \\ & = \sum _{j\in [m]} \frac {|S_r|^2 \ell ^{2d}}{(rn + j\ell )^d}\\ & \geq \frac {m|S_r|^2\ell ^{2d}}{(rn + m\ell )^d}, \end{align*}

Combining the two inequalities above gives us

\begin{align*} &\frac {m|S_r|^2\ell ^{2d}}{(rn + m\ell )^d} \leq \ell ^{2d} + rm|V|^{2r-1}\ell ^d\\ \implies & |S_r|^2 \leq \frac {(rn + m\ell )^d}{m} + r|V|^{2r-1}\frac {(rn + m\ell )^d}{\ell ^d}. \end{align*}

By setting $m = n^{\frac {d}{2rd + 1}}$ and $\ell = n^{1 -\frac {d}{2rd + 1} }$ , we get

\begin{equation*} \binom {|V|}{r}^2 = |S_r|^2 \leq cn^{d - \frac {d}{2rd + 1}} + c|V|^{2r-1}n^{\frac {d^2}{2rd + 1}}, \end{equation*}

for some constant $c$ depending only on $d$ and $r$ . We can solve from this inequality that

\begin{equation*} |V| = O\left (n^{\frac {d}{2r}\left (1 - \frac {1}{2rd + 1}\right )}\right ), \end{equation*}

completing the proof.