Why is the projection of a closed polytope closed? In general, projection of a closed set into a subspace does not result in a closed set. However, I was able to prove that in $\mathbb{R}^n$, the projection of a closed polytope (intersection of finitely many closed half spaces) onto a 1 dimensional subspace is closed. However the proof involved induction on dimension and using the theory of linear optimization.  
Is there a short topological proof of this fact? And perhaps a generalization that projection onto an m-dimensional subspace preserves closure? Intuitively imagining the shadow of a polytope, I feel it's true.
 A: This is a property of linear maps in $\mathbb{R}^n$, not just projections.
The key result here is that the class of polyhedral sets (intersection of a finite number of closed half-spaces) is the same as the class of finitely generated sets (Rockafellar, "Convex analysis", Theorem 19.1).
To quote Rockafellar, "This classical result is an outstanding example of a fact with is completely obvious to geometric intuition, but which wields important algebraic content and is not trivial to prove".
Suppose $P$ is a polyhedral set, then there are vectors $v_1,...,v_l, d_1,....,d_m$ ($l$ or $m$ may be zero with appropriate adjustments, the $v_i$ are 'points', the $d_j$ are 'directions') such that
$P = \{ \sum_i \lambda_i v_i + \sum_j \mu_j d_j | \sum_i \lambda_i =1, \lambda_i \ge 0, \mu_i \ge 0 \}$.
Suppose $A$ is a linear operator. Then
$AP = \{\sum_i \lambda_i A v_i + \sum_j \mu_j A d_j | \sum_i \lambda_i =1, \lambda_i \ge 0, \mu_i \ge 0 \}$, hence $AP$ is finitely generated, and so is a polyhedral set, hence closed.
A: There is a purely topological way to look at this. The other proofs are probably more sensible, but I wanted to rescue the intuition about limit points.  
First of all, if we're looking at a bounded polytope $P$, then $P$ is compact.  If $Y$ is any topological space, and $f:\mathbb{R}^n\to Y$ is any continuous function, then $f(P)$ is compact as well. If $Y$ is Hausdorff, this implies that $f(P)$ is closed.
This does not immediately work for an unbounded polytope, but it can be modified to work by compactification. For instance, if $P$ is a half-space with boundary containing the origin, then we can compactify $\mathbb{R}^n$ to a closed, solid ball — by adding a point at infinity to each ray from the origin — so that $P$ extends to a solid hemisphere. Then any projection onto a subspace $\mathbb{R}^n \to V$ extends naturally to a projection of balls, so we can apply the earlier compactness argument.
A: Suppose we have a closed polytope which has the same projection of the original solid and whose intersection with the projection space is also the projection (strictly, the projection of this intersection onto the projection space is the projection). Since the projection space takes the subspace topology, the intersection of this polytope with the projection space is also closed. The quotient topology on the projected space is given by taking cosets of the projection axis, which agrees with the topology given by the bijection sending a point in the projection space with its subspace topology to its coset by the projection axis.
Taking the minkowski sum of the polytope with the axis of projection, the union over the cosets of the polytope by the projection axis, would produce an appropriate closed set, but I don't have proof at hand that it's closed. So, you can convert the polytope to one with the same projection, which is closed, and whose intersection with the origin is identical as follows:

Suppose the polytope is bounded to one side of the projection surface.
Translating the polytope along the projection axis does not change its projection, by definition. Translate the solid to where it lies strictly to one side of the projection space. Reflecting the polytope across the projection axis, we see a vertex is a front vertex iff the line segment connecting it to its reflection passes through the interior to the polytope and/or its reflection.
Take the set of front vertices, and take as a translation vector the negative of one of their vector projections onto the projection axis. Translating the front vertices along the projection axis until they lie on the opposite side of the projection space from the rest of the solid, they can be separated from the back of the solid by the projection surface, and so upon all further translation we obtain a convex polytope from the intersection of the solid with the half-plane containing the back of the polytope with boundary the projection surface. The polytope's projection is unchanged by taking these variants, only its intersection with the projection surface.
In repeating this translation indefinitely, the intersection of the modified polytope with the projection surface converges to the projection: the edges which connect a vertex whose projection lies on the boundary of the projected solid to a point on the projection surface are truncations of the skewed edges of the original solid, and thus project to subsets of their projections. Hence, the vertices of the intersection become arbitrarily close to the projections of vertices on the boundary of the projection, and since the projections of edges from the polytope are edges themselves when nondegenerate, the agreement of the edges of the intersection with them is a function of the agreement of their vertices, and so on.

Otherwise, suppose the polytope is not bounded to one side of the projection surface.
The projection is the union of all intersections with the projection surface of translations the polytope along the projection axis. Therefore, the union of the projections of the two polytopes given by restricting the polytope to the half-spaces on either side of the projection surface is the projection of the union of these two polytopes - the original polytope. Then the previous case can be applied to determine that each half's projection is closed, which shows that their union is closed, which shows the original projection is closed.

I don't want to go into more detail than necessary, but you can construct polytopes from each vertex and its surroundings which have only that one vertex, a vertex's projection either lies on the boundary of or is in the interior of the projection, and is in the interior iff, moving to the vector space centered at the vertex, the origin of the analogous projection space is interior to the projection of the polytope constructed from the vertex's projection, which can be described in terms of the non-negative scalar multiples of the projections of its edges. I was working towards showing the projection was a convex polytope as well through this. I definitely prefer the approach copper.hat posted though, I don't find this definition of convex polytopes comfortable at all.
A: Projections are continuous functions, and continuous functions preserve limits. If a set contains all of its limit points, then the image of that set under a continuous function also contains all of its limit points. Not sure off the top of my head what the conditions are for this to hold in a general topological setting, but it sure should hold in Euclidean space.
