Let $A:=\{x_1, \ldots, x_{n+1}\}$ be an affinely independent. Then $\operatorname{int} (\operatorname{conv} A) \neq \emptyset$

Question

I'm trying to prove below result, taken from this answer. Could you have a check on my attempt?

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Let $V$ be a vector space. A finite set $\{x_1, \ldots, x_n\} \subset V$ is called affinely independent if $\lambda _1 x_1 +\ldots+\lambda_n x_n=0$ and $\lambda_1 +\ldots +\lambda_1=0$ imply $\lambda_1 =\cdots = \lambda_n=0$. Clearly, linear independence implies affine independence.

Theorem: Let $V$ be a $n$-dimensional Hausdorff topological vector space and $A:=\{x_1, \ldots, x_{n+1}\} \subset V$ affinely independent. Then $\operatorname{int} (\operatorname{conv} A) \neq \emptyset$.

Proof:

Let $v_1 := x_2-x_1, \ldots, v_n :=x_{n+1} - x_1$. Then $\{x_1, \ldots, x_{n+1}\}$ is affinely independent if and only if $\{v_1, \ldots, v_n\}$ is linearly independent. We have $\operatorname{conv} A =x_1+ \operatorname{conv} \{0, v_1, \ldots, v_n\}$. So it suffices to show that $\operatorname{int} (B) \neq \emptyset$ with $B := \operatorname{conv} \{0, v_1, \ldots, v_n\}$. We have $V$ is topologically isomorphic to $\mathbb R^n$, so wlog we assume $V=\mathbb R^n$. All norms on $V$ is equivalent, so we consider the norm $[\cdot]$ on $V$ defined by $$ [\lambda_1 v_1 +\cdots+\lambda_n v_n] := |\lambda_1|+ \cdots+|\lambda_n|. $$

Let $a := \frac{1}{2n} v_1 +\cdots \frac{1}{2n} v_n \in B$ and $r:=\frac{1}{4n}$. For $x \in \mathbb B(a, r)$, we have $$ \begin{align} [x-a] =&\left [\lambda_1 v_1 +\cdots+\lambda_n v_n - \left (\frac{1}{2n} v_1 +\cdots \frac{1}{2n} v_n \right) \right] \\ =& \left | \lambda_1- \frac{1}{2n} \right | + \cdots + \left | \lambda_n- \frac{1}{2n} \right | \end{align} $$

Then $\left | \lambda_i - \frac{1}{2n} \right | <\frac{1}{4n}$ and thus $\lambda_i \in \left [\frac{1}{4n}, \frac{3}{4n} \right]$ for all $i=1, \ldots, n$. Then $0 \le \lambda_1 +\cdots + \lambda_n \le \frac{3}{4} < 1$. Then $$ x = \left (1-(\lambda_1 +\cdots+\lambda_n)\right ) . 0 + \lambda_1 v_1 +\cdots+\lambda_n v_n \in B. $$

It follows that $\mathbb B(a, r) \subset B$ and $a \in \operatorname{int} (B)$. This completes the proof.

Answer 1

A new answer now to your new problem! Let a be in coA. Then a=$\lambda _{1}x_{1}+\lambda _{2}x_{2}+.....+\lambda _{n}x_{n}+\lambda _{n+1}x_{n+1}$ where $\sum_{1}^{n+1}\lambda _{i}=1$ and all the lambda's are nonnegative. The latter implies $\lambda_{1}=1-\lambda _{2}-....-\lambda _{n+1}$. Then a=$x_{1}+\lambda _{2}(x_{2}-x_{1})+\lambda _{3}(x_{3}-x_{1})+.....\lambda _{n+1}(x_{n+1}-x_{1})$. But $x_{2}-x_{1}, .....,x_{n+1}-x_{1}$ are n linearly independent vectors of V (because A is affinely independent) and since dimV=n they form a basis of V. Since all lambdas are nonnegative the set of all linear combinations of these n vectors is a convex cone C which contains n linearly independent elements and therefore intC nonempty. Notice that 0 is in the cone because we are allowed to set $\lambda_{1}=....=\lambda _{n+1}=0$ Now we have just shown that coA has the form $x_{1}$+C ie it is a translation of a convex cone with a nonempty interior and clearly coA itself has nonemty interior!!

Answer 2

You dont mention the dimension of V at the initial question! If the dimV=5 and your set A has 2 elements there is no way to have nonempty interior of the affine hull. Because that would mean that we can produce the space from one element! Do you mean relative interior?? Please explain!! Unless you assume that dimV=n (as in the Theorem)! Then by the definition of affine independent set the n vectors $x_{2}-x_{1},x_{3}-x_{1},....,x_{n+1}-x_{1}$ are linearly independent and since dimV=n, they produce the space. Let's assume that a is in affA. Then a=$\lambda _{1}x_{1}+\lambda _{2}x_{2}+.....+\lambda _{n}x_{n}+\lambda _{n+1}x_{n+1}$ where $\sum_{1}^{n+1}\lambda _{i}=1$.The latter implies $\lambda_{1}=1-\lambda _{2}-....-\lambda _{n+1}$. Then a=$x_{1}+\lambda _{2}(x_{2}-x_{1})+\lambda _{3}(x_{3}-x_{1})+.....\lambda _{n+1}(x_{n+1}-x_{1})$. But $x_{2}-x_{1}, .....,x_{n+1}-x_{1}$ are n linearly independent vectors of V (as explained above) and since dimV=n they form a basis of V. The lambda's are arbitrary and that implies that affA=$x_{1}$+V and also implies that affA contains the origin, so it is a subspace which contains n linearly independent elements (namely the n independent elements of A. Since $x_{1}+V$=V we obtain affA=V and therefore has nonempty interior!