Convex Hull preserves Subsets

Theorem

Let $X$ be a vector space over $\R$.

Let $U, V \subseteq X$ be non-empty with $U \subseteq V$.

Then:

$\map {\operatorname {conv} } U \subseteq \map {\operatorname {conv} } V$

where $\operatorname {conv}$ denotes the convex hull.

Proof

Let:

$u \in \map {\operatorname {conv} } U$

From the definition of the convex hull, there exists $u_1, u_2, \ldots, u_n \in U$ and $\lambda_1, \lambda_2, \ldots, \lambda_n \in \R_{> 0}$ with:

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

such that:

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

Since $U \subseteq V$, we have:

$u_i \in V$ for each $i$.

So, we have:

$u \in \map {\operatorname {conv} } V$

So:

if $u \in \map {\operatorname {conv} } U$ then $u \in \map {\operatorname {conv} } V$.

So, from the definition of subset, we have:

$\map {\operatorname {conv} } U \subseteq \map {\operatorname {conv} } V$

$\blacksquare$