Convex Hull is Smallest Convex Set containing Set

Theorem
Let $X$ be a vector space.

Let $U \subseteq X$.

Let $\map {\operatorname {conv} } U$ be the convex hull.

Then $\map {\operatorname {conv} } U$ is the smallest convex subset of $X$ containing $U$ in the sense that:


 * $\map {\operatorname {conv} } U$ is convex and if $K \subseteq X$ is a convex subset with $U \subseteq K$, we have that $\map {\operatorname {conv} } U \subseteq K$.

Proof
We have:


 * $\ds \map {\operatorname {conv} } U = \set {\sum_{j \mathop = 1}^n \lambda_j u_j : n \in \N, \, u_j \in U \text { and } \lambda_j \in \R_{> 0} \text { for each } j, \, \sum_{j \mathop = 1}^n \lambda_j = 1}$

Considering $n = 1$ and $\lambda_1 = 1$, we obtain:


 * $u \in \map {\operatorname {conv} } U$ for each $u \in U$.

Now we prove that $\map {\operatorname {conv} } U$ is convex.

Let $u, v \in \map {\operatorname {conv} } U$.

Then there exists $u_1, u_2, \ldots, u_n \in U$, $v_1, v_2, \ldots, v_m$ and $\lambda_1, \lambda_2, \ldots, \lambda_n, \mu_1, \ldots, \mu_m \in \R_{> 0}$ such that:


 * $\ds u = \sum_{i \mathop = 1}^n \lambda_i u_i$

and:


 * $\ds v = \sum_{j \mathop = 1}^n \mu_j v_j$

with:


 * $\ds \sum_{i \mathop = 1}^n \lambda_i = 1$

and:


 * $\ds \sum_{j \mathop = 1}^n \mu_j = 1$

Now let $t \in \closedint 0 1$.

We have:


 * $\ds t u + \paren {1 - t} v = \sum_{i \mathop = 1}^n t \lambda_i u_i + \sum_{j \mathop = 1}^m \paren {1 - t} \mu_j \lambda_j$

We have:

with $t \lambda_i \ge 0$ and $\paren {1 - t} \mu_j \ge 0$ for each $i, j$.

Since we have $u_i \in U$ for each $i$ and $v_j \in U$ for each $j$, we have:


 * $t u + \paren {1 - t} v \in \map {\mathrm {conv} } U$

So $\map {\mathrm {conv} } U$ is convex.