TITLE: Why is the projection of a closed polytope closed?
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.