Intersection of boundary facets of a simplicial complex Suppose you have an equidimensional $n$-dimensional simplicial complex $\Delta \subseteq \mathbb Q^n$; i.e., $\Delta$ is the union of finitely many $n$-simplices that intersect only along proper faces. (I really do mean to use the same $n$.)  By an $n$-simplex, I mean the convex hull of $n+1$ affinely independent points in $\mathbb Q^n$. In this setup, a boundary facet is any $(n-1)$-simplex that is a facet of exactly one of the $n$-simplices that make up $\Delta$.  Each boundary facet lies on a unique hyperplane, and the $n$-simplex to which it belongs lies entirely on one halfspace.
I'm having trouble proving the geometrically reasonable (maybe even obvious!) claim that the intersection of these halfspaces is contained in $\Delta$.
Question: Is it true that the intersection of these halfspaces is contained in $\Delta$?  If so, can you point me to a reference?  I couldn't find a proof of this in Ziegler's Lectures on Polytopes.
One thing that is throwing me off is that the claim isn't true if $\Delta$ isn't of full dimension.  For instance, consider the situation in this picture: counterexample picture.  Here the intersection of the boundary facets isn't even bounded.
I'm not a discrete mathematician, so thanks for bearing with me.
 A: I'll risk a proof, assuming that by "the intersection of these halfspaces" you mean
"the intersection of all the halfspaces of all the boundary facets."
Let point $x$ be in this intersection, 
and assume for the purposes of contradiction that $x$
is exterior to $\Delta$.
Form an arrangement of hyperplanes $\cal{A}$ by all the hyperplanes determined
by $x$ and all the ridges ($(n{-}2)$-dimensional facets, e.g., edges
of tetrahedra) of $\Delta$.
Find some point $z$ that is inside $\Delta$ but does not lie on $\cal A$.
Such a $z$ exists because $\Delta$ cannot lie entirely within a hyperplane
of $\cal{A}$.
Now aim a ray $r$ from $x$ to $z$, and
let $y$ be the first point on this
ray that coincides with a point of $\Delta$.
This $y$ must lie
in the relative interior of some
facet of some simplex $\sigma$ of $\Delta$,
for if instead it lay on a ridge, then $y$, and therefore $z$, would lie on $\cal{A}$.
The halfspace of this facet must be oriented so that it includes
$\sigma$ and excludes $x$.  So $x$ is afterall not inside the intersection of all
the boundary halfspaces.
Of course it is possible that the intersection of the halfspaces is the empty set,
but I assume you would treat that as contained in $\Delta$.
