# Showing that $\text{int}(\text{ch}(x_1, …, x_k))\subset\text{int}(\text{ch}(x_1, …, x_n))$

I am studying the book Convexity: An Analytic Viewpoint by B. Simon and I am not sure if understood a passage in the following proof (page 124):

Proposition 8.7 $\text{ch}(e_1, ..., e_n)$ always has a nonempty inteiror as a subset of $S(e_1, ..., e_n)$.

Proof $~~$ By successively throwing out dependent vectors from $P = \{e_j − e_1\}_{j=2}^n$, find a maximal independent subset of $P$. By relabeling, suppose it is $P' = \{e_j − e_1\}_{j=2}^k$ so $\{e_1, ... , e_k\}$ are affinely independent, and each $e_\ell − e_1$ with $\ell > k$ is a linear combination of $P'$. Then $\text{S}(e_1, . . . , e_n) = \text{S}(e_1, . . . , e_k )$.

$\color{red}{\text{Since}~\text{ch}(e_1, . . . , e_k ) \subset \text{ch}(e_1,...,e_n),~\text{it suffices to prove the result when}~e_1, . . . , e_n~\text{are}}$ $~\color{red}{\text{ affinely independent.}}$ In that case, $\varphi: \Delta_{n-1}\to \text{ch}(e_1, . . . , e_k )$ is a bijection and continuous, so a homeomorphism. Since $\Delta_{n-1}$ has a nonempty interior ($\{(\theta_1, . . . , \theta_n ) | \sum_{i=1}^n\theta_i = 1, 0 < \theta_i\}$), so does $\text{ch}(e_1, . . . , e_k )$.

So, if I didn't misunderstood nothing, we have to state the following:

Given points $x_1, ..., x_k, x_{k+1}, ..., x_n$ in a locally convex topological vector space, then $\text{int}(\text{ch}(x_1, ..., x_k))\subset\text{int}(\text{ch}(x_1, ..., x_n))$.

Here, "$\text{ch}(A)$" is the convex-hull and $\text{S}(A)$ is the affine span of $A$. The interior taken is relative to the subspace-topology of the affine span (the relative/intrinsic interior).

How can I state this result? At the end, I will need this result only for $\Bbb R^v$, for which the conclusion is geometrically clear.

Any hint to show it will be really appreciated.

Here $\subset$ must be interpreted as $\subseteq$ (i.e. equality is allowed), otherwise the statement is false.
$\text{ch}(x_1, \ldots, x_k) \subset \text{ch}(x_1, \ldots, x_n)$ is obvious (any convex combination of $x_1, \ldots, x_k$ is a convex combination of $x_1, \ldots, x_n$: just give coefficient $0$ to $x_{k+1}, \ldots, x_n$). If $A \subset B$ then $\text{int}(A) \subset \text{int}(B)$
• The key here is $A\subset B\Rightarrow \text{int}(A)\subset\text{int}(B)$, I just forgot this simple thing hold in general – Filburt Aug 2 '17 at 19:45