If $v_0\in\mathrm{hull}(v_0,\ldots,v_n)^\circ$ then $v_0\in\mathrm{hull}(v_1,\ldots,v_n)$ Let $v_0,\ldots,v_n\in\Bbb R^d$ be vectors, $\operatorname{hull}$ denote the convex hull, and $\operatorname{Int}$ denote the topological interior, then suppose $$v_0\in\operatorname{Int}(\operatorname{hull}(v_0,\ldots,v_n)).$$ It seems intuitive to me that we would then have that $$v_0\in\operatorname{hull}(v_1,\ldots,v_n)$$ but why is this true (assuming it is)?
 A: To quote Wikipedia (most reliable source there is xD):

the interior of a subset S of points of a topological space X consists of all points of S that do not belong to the boundary of S

So, it $v_0 \in Int(S)$, that means $v_0$ is in $S$ but it is not on the boundary os $S$. Specifically, $Int(S) \subset S \therefore \forall x \in Int(S), x \in S$. Your intuition is correct.
You can imagine this as some random polygon. The polygon contains all points strictly inside of it and all points on its border. The interior of the polygon contains all points strictly inside of it. If a point is an interior point of the polygon, then it is strictly inside of the polygon, and is in the polygon itself as well.
A: Without loss of generality, suppose that that the vectors are distinct.  Also without loss of generality (i.e. up to a translation of the convex hull), suppose that $v_0 = \vec 0$.  Now, note that if $0$ is in the interior of $\operatorname{hull}(v_0,v_1,\dots,v_n)$, then there exist $\lambda_i \geq 0$ with  $\sum \lambda_i = 1$ and $\lambda_0 \neq 1$ such that 
$$
\lambda_0 0 + \lambda_1 v_1 + \cdots + \lambda_n v_n = 0 \implies\\
\lambda_1 v_1 + \cdots + \lambda_n v_n = 0
$$
Now, let $\mu_i = \lambda_i\left(\sum_{i=1}^n \lambda_i\right)^{-1}$.  Then $\sum_i \mu_i = 1$, and 
$$
\mu_1 v_1 + \cdots + \mu_n v_n = \left(\sum_{i=1}^n \lambda_i\right)^{-1}(\lambda_1 v_1 + \cdots + \lambda_n v_n) = \left(\sum_{i=1}^n \lambda_i\right)^{-1}(0) = 0
$$
So, we see that $0 \in \operatorname{hull}(v_1,\dots,v_n)$, as desired.

To see that we can express $0$ as a convex combination with $\lambda_0 \neq 0$, note that: 


*

*$0$ is in the interior of the convex hull.  So, there is an open neighborhood about $0$ contained in the convex hull.  

*So, for a sufficiently small $\epsilon > 0$, $-\epsilon v_1$ is in the convex hull.  

*So, $0$ can be expressed as a convex combination of $-\epsilon v_1$ and $v_1$, which is in turn a convex combination of $v_0,v_1\dots,v_n$ with $\lambda_1 \neq 0$.

A: Hint: first you should assume $n \ge 1$ otherwise that's not true!
According to the definition of extreme points observe that
$$\rm ext ~( conv(v_0 , v_1, ..., v_n  ) ) \subseteq  \{ v_0 , v_1, ..., v_n \}  $$
Since $v_0$ is an interioir point then it cannot be an extreme point so
$$\rm ext ~( conv(v_0 , v_1, ..., v_n  ) ) \subseteq  \{   v_1, ..., v_n \}  $$
$$\rm conv~ [ ext ~( conv(v_0 , v_1, ..., v_n  ) )] \subseteq \rm conv \{ v_1, ..., v_n \}  $$
$$ \rm conv(v_0 , v_1, ..., v_n  )  \subseteq \rm conv \{ v_1, ..., v_n \}  $$
DONE.
This proof shows only if $v_0$ is not a extreme (vertex) point then claim still holds.
