From Shastri, Ch.2, though this is not the focus of the book.
The affine structure of $\mathbb{R}^d$ can be understood as follows. Given a finite subset $\{x_1, x_2, \cdots, x_n\}$ of $\mathbb{R}^d$, we say $\sum\lambda_ix_i$ is an affine combination of $x_i$’s if $\sum\lambda_i=0$.
The affine hull $\operatorname{aff}A$ of a subset $A\subset\mathbb{R}^d$ is then defined as all affine combinations in $A$. It is also clear how to define the notions of affine subspace and affine independent subsets.
It follows that
$\operatorname{aff}A$ is an affine subspace of $\mathbb{R}^d$, and is of course the smallest one containing $A$.
If $V\subset\mathbb{R}^d$ is an affine subspace, $V-x$ is a linear subspace for all $x\in A$. We then have a well-defined dimension $\dim V:=\dim(V-x)$.
$x_1, x_2,\cdots, x_n$ are affinely independent iff $x_2-x_1, x_3-x_1,\cdots, x_n-x_1$ are linearly independent. This implies $n\leq d+1$.
$\dim V+1$ is the size of any maximal affinely independent subset of $V$.
Theorem 1.(General Position) A subset $A\subset\mathbb{R}^d$ is said to be in general position if any $k$-subset of $A$ is affinely independent for $k\leq d+1$. Let $A=\{x_1, x_2, \cdots, x_n\}$. We have
Proof of 1. Induction on $n$. $n=1$ trivial. Suppose $y_1,\cdots,y_{n-1}$ is in general position. We need $y_n\notin$ some finite union of affine subspaces with $\dim\leq d-1$. The complement of this union is a finite intersection of open dense subsets, so is dense as well.
Proof of 2. Expanding this condition as a finite collection of $\det$ functions, we see this is an “open” condition.
Given a finite subset $\{x_1, x_2, \cdots, x_n\}$ of $\mathbb{R}^d$, we say $\sum\lambda_ix_i$ is an convex combination of $x_i$’s if $\sum\lambda_i=1$ and $\lambda_i\geq 0$ for all $i$. We then define the convex hull $\operatorname{conv}A$ and convex subsets.
Define a $n$-simplex as the convex hull of $n+1$ affinely independent points. These are of course (affinely) isomorphic to the standard $n$-simplex $\Delta_n$.
Theorem 2.(Carathéodory) Every $x\in\operatorname{conv}A$ lies in some $k$-simplex with vertices in $A$, where $k\leq d$.
Proof. Equivalently, we need $x$ is a convex combination of $d+1$ points in $A$. Suppose $x=\sum_{i=1}^{d+2}\lambda_ix_i$ is a convex combination and all $\lambda_i>0$.
As $x_1,\cdots,x_{d+2}$ are affinely dependent, we have $\sum\mu_ix_i=0$ with $\sum\mu_i=0$, and there is some $\mu_i>0$. We then find $\alpha=\operatorname{argmin}_i\{\lambda_i/\mu_i:\mu_i>0\}$. It follows that
\[x=\sum_{i=1}^{d+2}\lambda_ix_i=\sum_{i\neq\alpha}\left(\lambda_i-\mu_i\frac{\lambda_\alpha}{\mu_\alpha}\right)x_i,\]a convex combination of $d+1$ points.
Corollary 3. If $A$ is compact then $\operatorname{conv}A$ is compact.
Proof. We have shown the surjectivity of the map
\[A^{d+1}\times\Delta_d\longrightarrow\operatorname{conv}A.\]
It is also clear that if $A$ is convex then the closure $\overline A$ is also convex. Since, by continuity, the above map induces
\[\overline{A}^{d+1}\times\Delta_d\longrightarrow\overline{\operatorname{conv}A}=\overline{A},\]so $\operatorname{conv}\overline{A}\subset\overline{A}$.
We now introduce the notion of supporting half-spaces. Affine linear functions take the form $f(x)=\langle x,v\rangle+a$, where $\langle-,-\rangle$ is the standard inner product. Suppose $v\neq 0$, $H(f)=\{x:f(x)=0\}$ defines the hyperplanes and $H^+(f)=\{x:f(x)\geq 0\}$ defines the (closed) half-spaces.
Fix a convex subset $A$. $H^+$ is called a supporting half-space for $A$ if $A\subset H^+$. $H$ is called a supporting hyperplane if $H\cap A\neq\varnothing$ and $A\subset H^+$.
As a consequence,
Proposition 4. Every closed convex set in $\mathbb{R}^d$ is the intersection of its supporting half-spaces.
Definition. View a convex set $A$ as a subset of the affine subspace $\operatorname{aff}A$. Then we can talk about its relative interior $\operatorname{ri}A$ and relative boundary $\operatorname{rb}A$.
We can also define $\dim A:=\dim\operatorname{aff}A$.
Theorem 5. Let $A$ be a convex set in $\mathbb{R}^d$. Then every $x\in A\setminus\operatorname{ri}A$ is contained in some supporting hyperplane for $A$.
Proof. WLOG assume $x=0$. It suffices to prove the case $\operatorname{aff}A=\mathbb{R}^d$, as the supporting plane can be easily extended from $\operatorname{aff}A$ to $\mathbb{R}$. When $d=0$, $1$, there is nothing to prove. We will use induction for $d\geq 2$.
If $d=2$, consider the map (a composition of inclusion and quotient):
\[A\setminus 0\longrightarrow\mathbb{R}^2\setminus 0\longrightarrow S^1.\]Since $\dim A>1$, $A\setminus 0$ is connected, so the image in $S^1$ is an arc $\gamma$.
If the central angle of $\gamma$ is greater than $\pi$, then there is a triangle in $\gamma$, and hence $A$, with $0$ in its interior, contradicting the fact that $x\notin\operatorname{ri}A$. So there must be a line $0\in L\subset\mathbb{R}^2$ with $A\subset L^+$.
If $d>2$, consider a similar map
\[A\cap\mathbb{R}^2\setminus 0\longrightarrow\mathbb{R}^2\setminus 0\longrightarrow S^1,\]where $\mathbb{R}^2$ is regarded as a subset of $\mathbb{R}^d$. Note that $A’:=A\cap\mathbb{R}^2$ is an intersection of convex sets, hence convex. Two things can happen now:
- $\dim(A’)\leq 1$, or
- $\dim(A’)=2$. As before, $A’\setminus 0$ is connected and is mapped to an arc in $S^1$ with central angle $\leq\pi$.
In either case, there is a line $0\in L\subset\mathbb{R}^2$ such that $A’\subset L^+$. Construct the quotient
\[q:\mathbb{R}^d\longrightarrow\mathbb{R}^d/L\simeq\mathbb{R}^{d-1}.\]Since $q$ is linear, $q(A)$ is convex. We’ve shown that $0\notin\operatorname{int}q(A)$. So by induction hypothesis, there is some supporting hyperplane $H$ for $q(A)$ in $\mathbb{R}^d/L$. This clearly lifts to $q^{-1}(H)$, a supporting hyperplane for $A$.
Post date: 2025/02/21
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