Closed convex hull, Caratheodory Let $E$ be a compact set in $\mathbb{R}^{n}$ and let $x_{0}\in \overline{conv(E)}$. My goal is to show that $$x_{0}=\sum_{i=1}^{n+1}{\lambda_{i}x_{i}}$$ for $\lambda_{i}\geq 0$ with $\sum_{i=1}^{n+1}{\lambda_{i}}=1$ and $x_{i}\in E$. Using that  $conv(\overline{E})=\overline{conv(E)}$ (as $E$ is bounded) and Caratheodory's theorem, I can only deduce that $x_{0}=\sum_{i=1}^{n+1}{\lambda_{i}x_{i}}$ for $\lambda_{i}\geq 0$ and $x_{i}\in \overline{E}.$ Why is it possible to replace $\overline{E}$ by $E$ here?
Edit: I changed $E$ from being bounded to being compact.
 A: Edit: I think I see my misunderstanding. My bad (above "answer" erased)
Edit 2: I may have an actual solution now. Caratheodory's seems to give you $n$ points $x_i$ and $\lambda_i$ such that the convex combination is $x$. To see that these points can be taken within the interior of $E$, center coordinates about $x$ and scale in towards $x$ by a factor of $1-\epsilon$, lets say. This means move each $x_i$ to $(1-\epsilon)*x_i + \epsilon x$. This is within the closure of the convex hull (I think--see bottom), since the point itself is a convex combination, and if you compute 
$$
\sum \lambda_i((1-\epsilon)x_i + \epsilon x) = (1-\epsilon)\sum \lambda_i x_i + \epsilon \sum \lambda_i x = x
$$
However, we now must show that these new points are in $conv(E)$, not just the closure. First point. If $x_0$ is in the boundary, then we do not have any guarantees. In this case, I believe the theorem to be false. (Consider the cube with $x_0$ on a face--there is no way to write this as a convex combination of interior points. Just monitor the relevant coordinate.) 
If $x$ is in the interior, then there is some $\epsilon \in (0,1)$ so that the ball of radius $\epsilon$, centered at $x$ is entirely within $E$. We should take $\epsilon$ such that the new points are close enough to $x$ to be away from the boundary.
At this point I have my last worry: how to show that the new points are even in $E$? This might require some complex something or other, but at least we have no reduced the problem to a point (possible) on the boundary and a point $x$ that is in the interior. This means we have a 1-d problem instead of n.
Hope this is helpful!
A: This is not true, an easy counterexample is $E = (0,1)$.
