Proof. For $`v \in S`$, choose coefficients $`a_i`$, $`R`$ so that $`\{x \in \mathbb{R}^m \mathrel{:}\sum_{i=1}^m (x_i-a_i)^2 \le R \}`$ is a ball with $`v`$ on its boundary and with radius small enough so that no points of $`S \setminus v`$ are inside. By the Veronese map $`V^d_m`$, this ball corresponds to a hyperplane in $`\operatorname{\mathbb{R}}^{\binom{m+d}{m}-1}`$ containing $`v`$ and with all other points of $`S`$ on one side. This shows that $`V_m^d(v)`$ is a vertex of $`\operatorname{conv}(V_m^d(S))`$. For the second claim, let $`T \subset S`$, $`|T| \le \binom{m/2+d}{m/2}-1`$. Let $`p(x)=1`$ be a degree $`m/2`$ polynomial passing through each point of $`T`$ and no points of $`S \setminus T`$. To show that such a polynomial exists, recall we are assuming that $`V_{m/2}^d(S)`$ is in general position. Therefore, for any $`|T|`$ points in $`V_{m/2}^d(S)`$ there is a hyperplane passing through precisely those $`|T|`$ points. This hyperplane corresponds to a degree $`m/2`$ polynomial passing through each point of $`T`$ and no points of $`S \setminus T`$. Then $`\bigl(p(x)-1\bigr)^2= 0`$ is a polynomial surface which corresponds to a hyperplane in $`\operatorname{\mathbb{R}}^{\binom{m+d}{m}-1}`$ which supports $`\operatorname{conv}(T)`$ as a face of $`\operatorname{conv}(S)`$.  ◻

Proof. For $`v \in S`$, let $`H = \{x \in \operatorname{\mathbb{R}}^d : a\cdot x = 0\}`$ be a plane through the origin which contains $`v`$ and contains no other point of $`S`$. Then $`(a\cdot x)^m \le 0`$ is a degree $`m`$ homogeneous polynomial inequality which, by the homogeneous Veronese map $`HV^d_m`$, corresponds to a hyperplane in $`\operatorname{\mathbb{R}}^{\binom{m+d-1}{m}}`$ containing $`v`$ and with all other points of $`S`$ on one side. This shows that $`HV_{m}^d(v)`$ is a vertex of $`\operatorname{conv}(HV_{m}^d(S))`$. For the second claim, let $`T \subset S`$, $`|T| \le \left(\binom{m/2+d-1}{m/2}-1\right)`$. Let $`p(x)=0`$ be a degree $`m/2`$ homogeneous polynomial passing through each point of $`T`$ and no points of $`S \setminus T`$. To show that such a polynomial exists, recall we are assuming that $`HV_{m/2}^d(S)`$ is in general position. Therefore, for any $`|T|`$ points in $`HV_{m/2}^d(S)`$ there is a hyperplane passing through the origin and precisely those $`|T|`$ points. And this hyperplane corresponds to a degree $`m/2`$ homogeneous polynomial passing through each point of $`T`$ and no points of $`S \setminus T`$. Then $`(p(x))^2 = 0`$ is a degree $`m`$ homogeneous polynomial surface which corresponds to a hyperplane in $`\operatorname{\mathbb{R}}^{\binom{m+d-1}{m}}`$ which supports $`\operatorname{conv}(T)`$ as a face of $`\operatorname{conv}(S)`$.  ◻

Proof. Let $`v \in S`$. Choose a hyperplane $`H`$ containing $`v`$ and with all other points of $`S`$ on one side of it. Choose another hyperplane $`H'`$ parallel to $`H`$ and with all points of $`S`$ between $`H`$ and $`H'`$. Let $`S'`$ be the stereographic projection (using $`v`$ as the “pole”) of $`S \setminus v`$ onto $`H'`$. We claim that the number of $`k`$-facets of $`S`$ containing $`v`$ is equal to the number of $`k`$-facets of $`S'`$ (as a subset of $`H'`$, a $`(d-1)`$-dimensional subspace).

Assume that $`\operatorname{conv}(v_1 ,\dotsc , v_{d-1},v)`$ is a $`k`$-facet of $`S`$. We claim that for each $`s \in S`$, the stereographic projection $`s'`$ is on the positive side of $`\operatorname{aff}(v_1', \dotsc ,v_{d-1}')`$ if and only if $`s`$ is on the positive side of $`\operatorname{aff}(v_1, \dotsc , v_{d-1},v)`$. This is seen to be true by observing that $`\operatorname{aff}(s,v)`$ does not intersect $`\operatorname{aff}(v_1, \dotsc , v_{d-1},v)`$ anywhere other than the point $`v`$. This shows that $`\operatorname{conv}(v_1', \dotsc , v_{d-1}')`$ is a $`k`$-facet of $`S'`$. For the converse, assume that $`\operatorname{conv}(v_1', \dotsc ,v_{d-1}')`$ is a $`k`$-facet of $`S'`$. Then $`\operatorname{conv}(v_1, \dotsc ,v_{d-1},v)`$ is a $`k`$-facet of $`S`$ for the same reason as above.

Since $`S'`$ lies in a hyperplane, it can have at most $`e_k^{(d-1)}(n-1)`$ $`k`$-facets. Performing this projection on each point of $`S`$ and noticing that every $`k`$-facet is counted $`d`$ times shows the desired result. ◻

Proof. Consider the embedding $`\varphi:\operatorname{\mathbb{R}}^d \to \operatorname{\mathbb{R}}^{2k+d-1}`$ defined by

\varphi (x_1,x_2, \dotsc, x_d) = (x_1,x_1^2, x_1^3, \dotsc, x_1^{2k},x_2, \dotsc , x_d).

For each $`n \in \operatorname{\mathbb{N}}`$, let $`G_n \subset \operatorname{\mathbb{R}}^{dn}`$ consist of all configurations of $`n`$ points in $`\operatorname{\mathbb{R}}^d`$ such that no two points in the configuration have the same $`x_1`$-coordinate. One can verify that $`G_n`$ is open and dense in $`\operatorname{\mathbb{R}}^{dn}`$. For any $`n`$, let $`S \in G_n`$ be some configuration of $`n`$ points in $`G_n`$.

To show that $`\varphi(S)`$ is $`k`$-neighborly, let $`v_1, \dotsc , v_k`$ be $`k`$ points from $`S`$. Consider in the domain of the embedding, $`\mathbb{R}^d`$, the surface

\begin{equation}
\label{equ:surface}
\prod_{i=1}^k (x_1-v_{i1})^2 = 0.
\end{equation}

By expanding [equ:surface] we see that this surface corresponds via $`\varphi`$ to a hyperplane $`H`$ in $`\operatorname{\mathbb{R}}^{2k+d-1}`$. Note that $`v_1, \dotsc , v_k`$ satisfy [equ:surface] and all other points in $`S`$ satisfy $`\prod_{i=1}^k (x_1-v_{i1})^2 > 0`$. Using $`\varphi`$, we get that $`H`$ is a face-defining hyperplane that makes $`\varphi(v_1), \dotsc , \varphi(v_k)`$ a face of $`\operatorname{conv}(\varphi(S))`$. Therefore, $`\varphi(S)`$ is $`k`$-neighborly. This shows that $`p_g(k,d) \le 2k+d-1`$. ◻

Proof. If $`\varphi: \operatorname{\mathbb{R}}^d \to \operatorname{\mathbb{R}}^p`$ is generally $`k`$-neighborly, then for each $`n`$ the set of configurations of $`n`$ points which are mapped by $`\varphi`$ to $`k`$-neighborly point sets contains a set $`O`$ that is open and dense in $`\operatorname{\mathbb{R}}^{dn}`$. Let $`N \subset M^n`$ be the set of $`k`$-neighborly configurations of $`n`$-points in $`M^n`$. The set $`N`$ contains $`(\varphi \times \dotsb \times \varphi)( O)`$ which is open and dense since $`\varphi \times \dotsb \times \varphi`$ is a homeomorphism. Therefore $`M : = \varphi(\operatorname{\mathbb{R}}^d)`$ is a generally $`k`$-neighborly $`d`$-manifold. The other direction is similar.  ◻

Proof. Assume not, so that there exists some finite set $`S \subset M`$ and a set $`T`$ of $`k`$ points from $`S`$ such that no closed halfspace contains $`S`$ and contains $`T`$ on its boundary. This means that $`\operatorname{aff}(T) \cap \operatorname{relint}\operatorname{conv}(S \setminus T) \neq \emptyset`$ (from the separating hyperplane theorem ). We can pick $`|S|`$ small open balls in $`\operatorname{\mathbb{R}}^p`$ as follows. For each $`t \in T`$, let $`B_t`$ be a ball centered at $`t`$ and for each $`s \in S \setminus T`$, let $`A_s`$ be a ball centered at $`s`$. The radii of the balls can be chosen small enough so that any collection of points consisting of one point from each $`B_t`$ and one point from each $`A_s`$ has the property that the affine hull of the points from the $`B_t`$ intersects the relative interior of the convex hull of the points from the $`A_s`$. This means that any such configuration of points is not $`k`$-neighborly. Therefore, the subset of $`\operatorname{\mathbb{R}}^{p|S|}`$ of configurations of points of size $`|S|`$ on $`M`$ which are not $`k`$-neighborly contains $`\prod_{t \in T } (B_t \cap M) \times \prod_{s \in S \setminus T } (A_s \cap M)`$ which is an open subset of $`M^{|S|}`$. Therefore, the set of configurations of $`|S|`$ points on $`M`$ which are $`k`$-neighborly is not dense in $`M^{|S|}`$.  ◻

Proof. Fix $`k \geq 1`$. The “only if” direction is clear. We will now prove the “if” direction. Let $`T \subseteq S`$ be a set of $`k`$ points. Let $`\mathcal{U} = \{ U \subseteq S \mathrel{:}U \supseteq T \text{ and $U$ is finite}\}`$. For $`U \subseteq S`$ such that $`U \supseteq T`$, we will define $`N(U) \subseteq S^{p-1}`$ to be the set of unit outer normals to possible halfspaces $`H`$ such that $`T \subseteq \operatorname{bd}(H)`$ and $`U\subseteq H`$. More precisely, let

N(U) = \{a \in S^{p-1} \mathrel{:}
(\forall x,y \in T) a \cdot x = a \cdot y \text{ and } (\forall x \in T) (\forall y \in U) a \cdot x \geq a \cdot y \}.
%
%(\exists b \in \RR)[ (\forall x \in U) a \cdot x \leq b \text{ and } (\forall x \in S) a \cdot x = b]\}.

Clearly $`N(U)`$ is closed. Let $`\mathcal{V} \subseteq \mathcal{U}`$ be any finite subfamily. Then $`\cap_{U \in \mathcal{V}} N(U) = N(\cup_{U \in \mathcal{V}} U) \neq \emptyset`$ by assumption. We have established that $`\{N(U)\}_{U \in \mathcal{U}}`$ is a family of closed sets with the finite intersection property in compact space $`S^{p-1}`$. This implies $`\cap_{U \in \mathcal{U}} N(U) \neq \emptyset`$. We also have $`N(S) = N(\cup_{U \in \mathcal{U}} U) = \cap_{U \in \mathcal{U}} N(U) \neq \emptyset`$. That is, there is a halfspace $`H`$ such that $`T \subseteq \operatorname{bd}(H)`$ and $`S\subseteq H`$. As $`T \subseteq S`$ was arbitrary, this completes the proof.  ◻

Proof. Let $`P = \{ q_1, \dotsc, q_{p+2}\}`$. $`P`$ is affinely dependent and therefore there exist $`\lambda_1, \dotsc, \lambda_{p+2}`$ such that $`\sum_{i=1}^{p+2} \lambda_i q_i = 0`$, $`\sum_{i=1}^{p+2} \lambda_i = 0`$ and at least one $`\lambda_i`$ is non-zero. Because of the general position assumption, all $`\lambda_i`$ are non-zero. Let $`I = \{ i \mathrel{:}\lambda_i >0 \}`$, $`J = \{ i \mathrel{:}\lambda_i < 0\}`$. Both $`I`$ and $`J`$ are non-empty. By dividing $`\lambda_i`$s by $`\sum_{i \in I} \lambda_i`$ we can assume without loss of generality that $`\sum_{i \in I} \lambda_i = -\sum_{i \in J} \lambda_i = 1`$. Let $`Q = \{q_i \mathrel{:}i \in I \}`$, $`R = \{q_i \mathrel{:}i \in J \}`$. Let $`q = \sum_{i \in I} \lambda_i q_i \in \operatorname{relint}\operatorname{conv}Q`$, $`r = -\sum_{i \in J} \lambda_i q_i \in \operatorname{relint}\operatorname{conv}R`$. We have $`q=r`$, which completes the proof.  ◻

Proof. Assume $`Q,R`$ can be weakly separated. If the separation is not proper then $`\operatorname{aff}(Q \cup R) \neq \operatorname{\mathbb{R}}^p`$. If the separation is proper, then by the separating hyperplane theorem , $`\operatorname{relint}\operatorname{conv}Q \cap \operatorname{relint}\operatorname{conv}R = \emptyset`$.  ◻

Proof. First we will show that one can find arbitrarily large point sets in general linear position on $`C \setminus U`$. In order to accomplish this, first observe that $`C \setminus U`$ is non-degenerate. Indeed, if it were the case that $`C \setminus U`$ is contained in a hyperplane $`H`$, then we would have $`C = (C \cap H) \cup U`$ which is impossible since $`C`$ is irreducible.

Now assume that $`S`$ is a set of $`j`$ points in general linear position on $`C \setminus U`$. If it were not possible to find another point $`s`$ such that $`S \cup \{s\}`$ is in general linear position, it would have to be the case that $`C \setminus U`$ is contained in the union of all hyperplanes spanned by $`p`$ points from the set $`S`$. Let $`\mathcal{H}`$ be the collection of all hyperplanes spanned by $`p`$ points in $`S`$. Note that $`\mathcal{H}`$ is finite. We have

C = \bigg(\bigcup_{H \in \mathcal{H}} \bigl((C \setminus U) \cap H\bigr) \bigg) \cup U.

Since $`C \setminus U`$ is non-degenerate the above formula would be a representation of $`C`$ as the union of proper subvarieties. This is impossible since $`C`$ is irreducible. Therefore we know that $`C \setminus U`$ is not contained in the union of all hyperplanes spanned by points in $`S`$ and so we can always find $`s`$ so that $`S \cup \{s\}`$ is in general linear position. It follows that we can find arbitrarily large point sets in general linear position on $`C \setminus U`$.

Let $`P`$ be a set of $`p+2`$ points in general linear position on $`C \setminus U`$. By [lem:radon,lem:separation], there is a partition of $`P`$ into non-empty sets $`Q`$ and $`R`$ so that $`Q`$ and $`R`$ cannot be weakly separated.

However, because $`C \setminus U`$ is weakly $`k`$-neighborly, we know that for any set $`T`$ of $`k`$ points on $`C\setminus U`$ there exists a closed halfspace which contains $`C\setminus U`$ and contains $`T`$ in its boundary. In other words, any such $`T`$ can be weakly separated from $`C\setminus U`$. Therefore, it must be that $`k< \min(|Q|,|R|)`$, i.e. $`k \le \min(|Q|,|R|)-1`$. Now since $`\min(|Q|,|R|) \le \lfloor\frac{p+2}{2}\rfloor`$, we have that $`k \le \lfloor\frac{p+2}{2}\rfloor - 1`$ and so $`p \ge 2k`$. ◻

Proof. We identify the set of hyperplanes in $`\operatorname{\mathbb{R}}^p`$ with a proper subset of $`\mathbb{P}^p(\operatorname{\mathbb{R}})`$. This identification works as follows. We identify a hyperplane $`a_0 +a_1x_1 + \dotsb + a_px_p=0`$ in $`\operatorname{\mathbb{R}}^p`$ with the point with homogeneous coordinates $`(a_0,a_1, \dotsc , a_p)`$ in $`\mathbb{P}^p(\operatorname{\mathbb{R}})`$. Therefore, the set of hyperplanes in $`\operatorname{\mathbb{R}}^p`$ is identified with $`\mathbb{P}^p(\operatorname{\mathbb{R}}) \setminus \{(a_0, a_1, a_2, \dotsc , a_p) \mathrel{:}a_1 = a_2 = \cdots =a_p =0\}`$.

Recall from 16.3.1 that around any smooth real point of $`V`$ there is a neighborhood which is a smooth $`d`$-dimensional submanifold of $`\operatorname{\mathbb{R}}^p`$. Let $`T \subset \mathbb{P}^p(\operatorname{\mathbb{R}})`$ be the set of hyperplanes $`H`$ for which there exists a smooth point of $`V \setminus U`$ and a neighborhood of that smooth point which has nonempty transversal intersection with $`H`$. We know that $`T`$ is open and that for any $`H \in T`$, $`H \cap V`$ contains an open subset which is a smooth $`(d-1)`$-dimensional submanifold of $`\operatorname{\mathbb{R}}^p`$ . Let $`H`$ be any hyperplane in $`T`$. Clearly the dimension of $`H \cap V`$ is at most $`d-1`$. We claim that $`H \cap V`$ is a variety of dimension precisely $`d-1`$ and that it contains a smooth real point. Recall from 16.3.1 that if a variety has real dimension $`d`$ then it has dimension at least $`d`$. Therefore, the dimension of $`H \cap V`$ is $`d-1`$. Now we show that $`H \cap V`$ contains a smooth real point. Indeed assume not, so that $`H \cap V(\operatorname{\mathbb{C}})`$ contains no smooth real points. ³ This means that $`H \cap V`$ is contained in the set of singular points of the $`(d-1)`$-dimensional variety $`H \cap V(\operatorname{\mathbb{C}})`$. By , the set of singular points of $`H \cap V(\operatorname{\mathbb{C}})`$ is a variety of dimension at most $`d-2`$. Since $`H \cap V`$ has real dimension $`d-1`$ and is contained in a variety of dimension at most $`d-2`$ this is a contradiction. So $`H \cap V`$ contains smooth real points.

Let $`I`$ be the subset of $`T`$ consisting of hyperplanes $`H`$ such that $`H \cap V`$ is irreducible and non-degenerate. We will show that $`I`$ is open and dense in the standard topology in $`T`$. Let $`\overline{V(\operatorname{\mathbb{C}})} \subset \mathbb{P}^p(\operatorname{\mathbb{C}})`$ be the projective closure of $`V(\operatorname{\mathbb{C}})`$. This means that $`V(\operatorname{\mathbb{C}}) = \overline{V(\operatorname{\mathbb{C}})}\cap \{x_0 \neq 0\}`$. Because $`V`$ contains a smooth real point, $`V`$ is Zariski dense in $`V(\operatorname{\mathbb{C}})`$, that is, $`V(\operatorname{\mathbb{C}})`$ is the smallest complex variety containing $`V`$, see  [[alternatively, Theorem 5.1 in https://arxiv.org/pdf/1606.03127.pdf ]] . Therefore, since $`V`$ is irreducible, by , $`V(\operatorname{\mathbb{C}})`$ is irreducible. Because $`V(\operatorname{\mathbb{C}})`$ is irreducible, it is then a standard fact that $`\overline{V(\operatorname{\mathbb{C}})}`$ is irreducible. By projective duality, we can identify the set of all projective hyperplanes with real coefficients in $`\mathbb{P}^p(\operatorname{\mathbb{C}})`$ with $`\mathbb{P}^p(\operatorname{\mathbb{R}})`$. Let $`I' \subset \mathbb{P}^p(\operatorname{\mathbb{R}})`$ consist of all projective hyperplanes with real coefficients that have irreducible and non-degenerate intersection with $`\overline{V(\operatorname{\mathbb{C}})}`$. By , $`I'`$ is Zariski open and dense. ⁴

We will now show that the fact that $`I'`$ is Zariski open and dense implies that $`I`$ is also Zariski open and dense in $`T`$. Given a hyperplane $`H`$ defined by $`a_0 + a_1x_1 + \dotsb + a_px_p=0`$ in $`\operatorname{\mathbb{R}}^p`$, there is a corresponding projective hyperplane $`\overline{H}`$ defined by $`a_0x_0 + a_1x_1 + \dotsb + a_px_p=0`$ which is the homogenization of $`H`$. Let $`H`$ be a hyperplane in $`T`$ and assume that the homogenization $`\overline{H}`$ is in $`I'`$. We claim that this implies that $`H \in I`$. To establish this claim, we need to verify that $`H \cap V`$ is irreducible and nondegenerate. Observe that $`H \cap V(\operatorname{\mathbb{C}}) = \overline{H} \cap \overline{V(\operatorname{\mathbb{C}})} \cap \{x_0 \neq 0\}`$ i.e., $`H \cap V(\operatorname{\mathbb{C}})`$ is an open subset of $`\overline{H} \cap \overline{V(\operatorname{\mathbb{C}})}`$. It is a standard fact that a nonempty open subset of an irreducible space is irreducible and dense. Therefore $`H \cap V(\operatorname{\mathbb{C}})`$ is irreducible. Now, since $`H \cap V(\operatorname{\mathbb{C}})`$ is Zariski dense in $`\overline{H} \cap \overline{V(\operatorname{\mathbb{C}})}`$, if $`H \cap V(\operatorname{\mathbb{C}})`$ were contained in a hyperplane, $`\overline{H} \cap \overline{V(\operatorname{\mathbb{C}})}`$ would be as well. So $`H \cap V(\operatorname{\mathbb{C}})`$ is nondegenerate. We have shown that $`H \cap V(\operatorname{\mathbb{C}})`$ is non-degenerate and irreducible. We claim that because the section $`H \cap V(\operatorname{\mathbb{C}})`$ contains smooth real points, then the non-degeneracy and irreducibility of $`H \cap V(\operatorname{\mathbb{C}})`$ implies non-degeneracy and irreducibility of $`H \cap V`$. This relies on the fact mentioned above that if an irreducible variety $`W`$ contains a smooth real point, then the set of real points is Zariski dense in $`W(\operatorname{\mathbb{C}})`$. This means that $`H \cap V`$ is Zariski dense in $`H \cap V(\operatorname{\mathbb{C}})`$. Now by , irreducibility of $`H \cap V(\operatorname{\mathbb{C}})`$ implies that $`H \cap V`$ is irreducible. Finally, observe that because $`H \cap V`$ is Zariski dense in $`H \cap V(\operatorname{\mathbb{C}})`$, if $`H \cap V`$ is degenerate, $`H \cap V(\operatorname{\mathbb{C}})`$ must be as well. Thus $`H \cap V`$ is non-degenerate.

Recall that $`T`$ is open (in the standard metric topology) and that $`I'`$ is Zariski open and dense in $`\mathbb{P}^p(\operatorname{\mathbb{R}})`$ which implies that $`I'`$ is open and dense in the standard topology. We established that $`I`$ contains $`I' \cap T`$. Therefore, we have shown that $`I`$ is open and dense in the standard topology in $`T`$.

Let $`O \subset \mathbb{P}^p(\operatorname{\mathbb{R}})`$ be the set of hyperplanes which intersect $`B`$. The set $`O`$ is open (in the standard topology).

We claim that all of this means that $`I \cap T \cap O`$ is non-empty. To establish this, we need to verify that $`T \cap O`$ is non-empty. To find $`H \in T \cap O`$, let $`v`$ be a smooth point of $`V \setminus U`$ and let $`F`$ be a $`(p-2)`$-flat having non-empty transversal intersection with a neighborhood of $`v`$. Such a flat exists because $`V`$ has real dimension at least 2. For any point $`p`$ in $`B`$, the hyperplane $`\operatorname{aff}(F \cup p)`$ is in $`T \cap O`$.

Therefore, $`T \cap O`$ is open and non-empty. Since $`I`$ is open and dense in $`T`$, it follows that $`I \cap T \cap O \neq \emptyset`$. We claim that any $`H \in I \cap T \cap O`$ completes the proof. To show this, it remains to show that for any $`H \in I \cap T \cap O`$, $`H \cap V`$ has a smooth real point which is not in $`U`$. We already know that $`H \cap V`$ has smooth real points and that $`H \cap V`$ is not contained in $`U`$. So assume for a contradiction that all the smooth real points are contained in $`U`$. Then letting $`S`$ denote the singular points of $`H\cap V`$, we have that $`H \cap V = (H \cap V \cap U)\cup S`$ is the decomposition of $`H \cap V`$ as the union of two proper closed subvarieties, a contradiction to irreducibility of $`H\cap V`$. ◻

Proof of 51. First we show that by making repeated applications of 53, we can inductively construct a $`(p-d+1)`$-flat $`L`$ that intersects $`\operatorname{int}\operatorname{conv}(V \setminus U)`$ and such that $`L \cap V`$ is a non-degenerate irreducible algebraic curve with a smooth real point which is not contained in $`U`$. Say we have some flat $`F`$ that intersects $`\operatorname{int}\operatorname{conv}(V \setminus U)`$ and such that $`F \cap V`$ is a non-degenerate irreducible $`d'`$-dimensional $`(d' \ge 2)`$ variety with a smooth real point not contained in $`U`$. Since $`F \cap V`$ is not contained in $`U`$, $`F \cap U`$ is a proper subvariety of $`F \cap V`$. Let $`B`$ be an open ball in $`F`$ that is contained in $`\operatorname{int}\operatorname{conv}(V \setminus U)`$. By 53, there exists a hyperplane $`H`$ in $`F`$ that intersects $`B`$ and such that $`H \cap V`$ is a non-degenerate irreducible $`(d'-1)`$-dimensional variety with a smooth real point not contained in $`F \cap U`$ and hence not contained in $`U`$. Since $`B`$ is contained in $`\operatorname{int}\operatorname{conv}(V \setminus U)`$, $`H`$ intersects $`\operatorname{int}\operatorname{conv}(V \setminus U)`$. We can repeat this process until we obtain a $`(p-d+1)`$-flat $`L`$ that intersects $`\operatorname{int}\operatorname{conv}(V \setminus U)`$ and such that $`C:=L \cap V`$ is a non-degenerate irreducible algebraic curve with a smooth real point not contained in $`U`$.

Now we will show that $`C \setminus U`$ is weakly $`k`$-neighborly in $`L`$, meaning⁵ that $`C \setminus U`$ is weakly $`k`$-neighborly as a subset of its affine hull $`L`$.

Because $`V \setminus U`$ is weakly $`k`$-neighborly, we know that for any set $`T`$ of $`k`$ points on $`C\setminus U`$ there exists a closed halfspace $`H`$ (in $`\operatorname{\mathbb{R}}^p`$) which contains $`C \setminus U`$ and contains $`T`$ in $`\operatorname{bd}(H)`$. In other words, any such $`T`$ can be weakly separated from $`C \setminus U`$. Notice that we are talking about weak separation in $`\operatorname{\mathbb{R}}^p`$, while we are really interested in weak separation in $`L`$. We claim that our assumption that $`L`$ intersects the interior of the convex hull of $`V \setminus U`$ allows us to pass from weakly separating hyperplanes in $`\operatorname{\mathbb{R}}^p`$ to weakly separating hyperplanes in $`L`$. Indeed, the fact that $`L`$ intersects $`\operatorname{int}\operatorname{conv}(V \setminus U)`$ means that any closed halfspace $`H`$ satisfying $`V \setminus U \subset H`$ cannot contain $`L`$ in its boundary. Therefore $`\operatorname{bd}(H) \cap L`$ is a proper hyperplane in $`L`$. To summarize, given a set $`T`$ of $`k`$ points on $`C\setminus U`$, by the assumption that any finite set on $`V \setminus U`$ is weakly $`k`$-neighborly, there exists a hyperplane $`\operatorname{bd}(H)`$ that weakly separates $`T`$ from $`V \setminus U`$. This and the way we chose $`L`$ allows us to conclude that $`\operatorname{bd}(H) \cap L`$ is a hyperplane in $`L`$ that weakly separates $`T`$ from $`C\setminus U`$. Therefore, $`C \setminus U`$ is weakly $`k`$-neighborly in $`L`$. Since $`C`$ is not contained in $`U`$, $`C \cap U`$ is a proper subvariety. By 52, $`p-d+1 \ge 2k`$. ◻

Proof of 41. We can without loss of generality assume that $`V`$ is non-degenerate since otherwise we could consider $`V`$ in $`\operatorname{aff}(V)`$. By 47, every finite set of points on $`V_{\text{sm}}`$ is weakly $`k`$-neighborly. Therefore by 48, $`V_{\text{sm}}`$ is weakly $`k`$-neighborly. Since $`V_{\text{sing}}`$ is a proper closed subvariety (see ), by 51, $`p \ge 2k+d-1`$. ◻

Proof. Assume not, so that $`S`$ is a set of $`2k`$ points in $`M`$ such that $`\operatorname{aff}(S)\cap M`$ is not contained in the union of all hyperplanes supporting facets of $`\operatorname{conv}(S)`$. Then identifying $`\operatorname{aff}(S)`$ with $`\operatorname{\mathbb{R}}^{2k-1}`$, we can find a set $`S'`$ of $`2k+1`$ points in general position in $`\operatorname{\mathbb{R}}^{2k-1}`$ which are weakly $`k`$-neighborly. However, by [lem:radon,lem:separation], there is a partition $`Q,R`$ of $`S'`$ into two non-empty sets so that $`Q,R`$ cannot be weakly separated. Since $`\min(Q,R) \le \lfloor \frac{2k+1}{2} \rfloor= k`$, this is a contradiction to weakly $`k`$-neighborliness of $`S'`$.  ◻