[proofplan] We define $S$ to be the set of all finite convex combinations of points of $A$. First we prove that $S$ is convex and contains $A$, so the convex hull, being the smallest convex set containing $A$, is contained in $S$. Then we prove that every convex subset of $\mathbb{R}^n$ containing $A$ contains $S$, by induction on the number of points in the finite convex combination. Applying this to $\operatorname{conv}(A)$ gives the reverse inclusion. [/proofplan] [step:Define the set of finite convex combinations] Define $S \subset \mathbb{R}^n$ by \begin{align*} S = \left\{ \sum_{i=1}^{m} \lambda_i x_i : m \in \mathbb{N},\ x_i \in A,\ \lambda_i \geq 0\ \text{for each } i,\ \sum_{i=1}^{m} \lambda_i = 1 \right\}. \end{align*} We will prove $\operatorname{conv}(A) = S$. [/step] [step:Show that $S$ is a convex set containing $A$] If $a \in A$, then \begin{align*} a = 1 \cdot a, \end{align*} so $a \in S$ by taking $m = 1$, $x_1 = a$, and $\lambda_1 = 1$. Hence $A \subset S$. Now let $y,z \in S$ and let $t \in [0,1]$. By definition of $S$, there exist $m,k \in \mathbb{N}$, points $x_1,\dots,x_m \in A$, points $w_1,\dots,w_k \in A$, coefficients $\lambda_1,\dots,\lambda_m \geq 0$, and coefficients $\mu_1,\dots,\mu_k \geq 0$ such that \begin{align*} y &= \sum_{i=1}^{m} \lambda_i x_i, & z &= \sum_{j=1}^{k} \mu_j w_j, & \sum_{i=1}^{m} \lambda_i &= 1, & \sum_{j=1}^{k} \mu_j &= 1. \end{align*} Then \begin{align*} t y + (1-t) z = \sum_{i=1}^{m} t\lambda_i x_i + \sum_{j=1}^{k} (1-t)\mu_j w_j. \end{align*} The coefficients $t\lambda_i$ and $(1-t)\mu_j$ are nonnegative, and their sum is \begin{align*} \sum_{i=1}^{m} t\lambda_i + \sum_{j=1}^{k} (1-t)\mu_j = t \sum_{i=1}^{m} \lambda_i + (1-t)\sum_{j=1}^{k} \mu_j = t + (1-t) = 1. \end{align*} Thus $t y + (1-t)z$ is a finite convex combination of points of $A$, so $t y + (1-t)z \in S$. Therefore $S$ is convex. [/step] [step:Deduce that the convex hull is contained in $S$] Since $S$ is convex and $A \subset S$, the defining minimality of the convex hull gives \begin{align*} \operatorname{conv}(A) \subset S. \end{align*} [/step] [step:Show that every convex set containing $A$ contains all elements of $S$] Let $C \subset \mathbb{R}^n$ be a convex set such that $A \subset C$. We prove that $S \subset C$. Let $m \in \mathbb{N}$, let $x_1,\dots,x_m \in A$, and let $\lambda_1,\dots,\lambda_m \geq 0$ satisfy \begin{align*} \sum_{i=1}^{m} \lambda_i = 1. \end{align*} We prove by induction on $m$ that \begin{align*} \sum_{i=1}^{m} \lambda_i x_i \in C. \end{align*} For $m=1$, the condition gives $\lambda_1 = 1$, so \begin{align*} \sum_{i=1}^{1} \lambda_i x_i = x_1 \in A \subset C. \end{align*} Assume the claim holds for convex combinations of $m-1$ points, where $m \geq 2$. If $\lambda_m = 1$, then $\lambda_i = 0$ for $1 \leq i \leq m-1$, and \begin{align*} \sum_{i=1}^{m} \lambda_i x_i = x_m \in A \subset C. \end{align*} If $\lambda_m < 1$, define \begin{align*} \alpha &= 1-\lambda_m, & \nu_i &= \frac{\lambda_i}{\alpha} \quad \text{for } 1 \leq i \leq m-1. \end{align*} Then $\alpha > 0$, each $\nu_i \geq 0$, and \begin{align*} \sum_{i=1}^{m-1} \nu_i = \frac{1}{\alpha}\sum_{i=1}^{m-1}\lambda_i = \frac{1-\lambda_m}{1-\lambda_m} = 1. \end{align*} By the induction hypothesis, \begin{align*} u := \sum_{i=1}^{m-1} \nu_i x_i \in C. \end{align*} Since $x_m \in A \subset C$ and $C$ is convex, the point \begin{align*} \alpha u + \lambda_m x_m \end{align*} belongs to $C$. Expanding this expression gives \begin{align*} \alpha u + \lambda_m x_m = \alpha \sum_{i=1}^{m-1} \nu_i x_i + \lambda_m x_m = \sum_{i=1}^{m-1} \lambda_i x_i + \lambda_m x_m = \sum_{i=1}^{m} \lambda_i x_i. \end{align*} Thus every finite convex combination of points of $A$ belongs to $C$, so $S \subset C$. [guided] The point of this step is to justify the phrase “convex sets contain convex combinations” for arbitrary finite combinations, not only for combinations of two points. Let $C \subset \mathbb{R}^n$ be convex and suppose $A \subset C$. We prove that every element of $S$ lies in $C$. Choose $m \in \mathbb{N}$, points $x_1,\dots,x_m \in A$, and coefficients $\lambda_1,\dots,\lambda_m \geq 0$ satisfying \begin{align*} \sum_{i=1}^{m} \lambda_i = 1. \end{align*} We must prove \begin{align*} \sum_{i=1}^{m} \lambda_i x_i \in C. \end{align*} We argue by induction on $m$. For $m=1$, the coefficient condition forces $\lambda_1=1$, hence \begin{align*} \sum_{i=1}^{1}\lambda_i x_i = x_1. \end{align*} Since $x_1 \in A$ and $A \subset C$, this point lies in $C$. Now assume the statement has been proved for convex combinations of $m-1$ points, with $m \geq 2$. There are two cases. If $\lambda_m=1$, then the sum of all coefficients is already $1$, so every earlier coefficient must be $0$. Therefore \begin{align*} \sum_{i=1}^{m}\lambda_i x_i = x_m \in A \subset C. \end{align*} It remains to handle the case $\lambda_m < 1$. We separate the last point from the first $m-1$ points. Define \begin{align*} \alpha &= 1-\lambda_m, & \nu_i &= \frac{\lambda_i}{\alpha} \quad \text{for } 1 \leq i \leq m-1. \end{align*} Because $\lambda_m < 1$, we have $\alpha > 0$, so the numbers $\nu_i$ are well-defined and nonnegative. They also sum to $1$: \begin{align*} \sum_{i=1}^{m-1}\nu_i = \frac{1}{\alpha}\sum_{i=1}^{m-1}\lambda_i = \frac{1-\lambda_m}{1-\lambda_m} = 1. \end{align*} Thus the point \begin{align*} u := \sum_{i=1}^{m-1}\nu_i x_i \end{align*} is a convex combination of $m-1$ points of $A$. By the induction hypothesis, $u \in C$. Also $x_m \in A \subset C$. Since $C$ is convex and $\alpha,\lambda_m \geq 0$ with $\alpha+\lambda_m=1$, we get \begin{align*} \alpha u + \lambda_m x_m \in C. \end{align*} Finally, \begin{align*} \alpha u + \lambda_m x_m = \alpha \sum_{i=1}^{m-1}\nu_i x_i + \lambda_m x_m = \sum_{i=1}^{m-1}\lambda_i x_i + \lambda_m x_m = \sum_{i=1}^{m}\lambda_i x_i. \end{align*} Therefore the original finite convex combination lies in $C$. Since the finite convex combination was arbitrary, $S \subset C$. [/guided] [/step] [step:Apply the universal property to the convex hull] The set $\operatorname{conv}(A)$ is convex and contains $A$. Applying the previous step with $C = \operatorname{conv}(A)$ gives \begin{align*} S \subset \operatorname{conv}(A). \end{align*} Together with $\operatorname{conv}(A) \subset S$, this proves \begin{align*} \operatorname{conv}(A) = S. \end{align*} [/step]