The set of all convex combinations of $n+1$ vectors from $A$ is the convex hull of $A$ if $A \subseteq \mathbb{R}^n$

Question

Let $A\subseteq \mathbb{R}^n$. If we mark:

$$A_k=\{\sum^k_{i=1}\lambda_i a_i: \sum^k_{i=1} \lambda_i=1,a_i\in A, \lambda_i \geq 0 \}$$

Then we want to show that:

$$A_{n+1}=C(A)$$

Where $C(A)$ is the convex hull of $A$.

Answer 1

It is known that the convex hull is the same as the set of all convex combinations. In other words:

$$\bigcup_{i=1}^{\infty} A_i = C(A)$$

In addition it is obvious that for all $k$ we have $A_k \subseteq A_{k+1}$ because we can always set $\lambda_{k+1}=0$. We will now prove that if $k>n+1$ then $A_k\subseteq A_{k-1}$.

Let $v\in A_k$. In other words we have some convex combinations s.t. $$\sum_{i=1}^{k}\lambda_ia_i=v,\ \sum_{i=1}^k\lambda_i=1, \ \lambda_i\geq1$$ We can express the first two conditions using the following linear equation:

$$ \begin{pmatrix} | & & | \\ a_1 & \cdots & a_k \\ | & & | \\ 1 & \cdots & 1 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ \vdots \\ x_k \end{pmatrix} = \begin{pmatrix} v_1 \\ \vdots \\ v_{n} \\ 1 \end{pmatrix}$$ Then the above system of linear equations has a solution $\lambda$ s.t. $\lambda_i\geq0$. Now, let $\Lambda$ be the solution space of the above linear equation. Said solution space is path connected. Now notice that the system of linear equations above has $k$ variables and $n+1<k$ equations. Thus because it has a solution it has at least one degree of freedom. That means we can create another solution $\Delta$ that has $\Delta_i < 0$ for some coordinate. Because $\Lambda$ is path connected we have a path $\gamma$ s.t. $$\gamma(0)=\lambda, \ \gamma(1)=\Delta$$ Now build the continuous function: $$f(v)=\min\{v_1, \cdots, v_k\}$$ And now $f(\gamma(0)) \geq 0$ and $f(\gamma(1)) < 0$ Thus exists $f(\gamma(t))=0$ but then $s=\gamma(t)\in \Lambda$ is a solution to the system of linear equations above s.t. $s_i\geq 0$ and exists $s_j=0$. This means that $v$ can be built as a convex combination of $k-1$ vectors using the values of $s$ as coefficients ignoring the one vector that has $s_j=0$.

Finally:

$$C(A) = \bigcup_{i=1}^{\infty} A_i = \bigcup_{i=1}^{n+1} A_i = A_{n+1}$$