Is the $n-2$ skeleton of a polytope closed in the Zariski topology?

Question

This question depends on the belief/assumption that hyperplanes are a special type of algebraic variety. If they are not, please correct me, and the question should end here.

Is the $n-2$ skeleton of a convex polytope closed in the Zariski topology?

This question is to some extent proof verification since I already have putative proof in my mind.

Attempted proof:

Every convex polytope can be written as the intersection of finitely many half-spaces. (See, e.g., here.) In particular, there always exists a unique minimal description using the facet-defining halfspaces (facet = $n-1$ face).

I believe then that each $n-2$ face of the convex polytope will then as a result be the intersection of some subset of the hyperplanes corresponding to the facet-defining halfspaces (the hyperplanes themselves are $n-1$ dimensional).

If each of the hyperplanes is closed in the Zariski topology, then so should be all of their intersections, which include the $n-2$ faces of the convex polytope.

The $n-2$ skeleton is the union of finitely many $n-2$ faces, so if each $n-2$ face is closed in the Zariski topology, then so should be the $n-2$ skeleton.

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This question is a follow-up to two of my previous questions:

What is the equation of the orthogonal group (as a variety/manifold)?

Why is the intersection of algebraic subsets of an algebraic variety again an algebraic subset?

Answer 1

One gap is in the third paragraph of your proof. The faces are not intersection of hyperplanes! So you can not conclude from closness of hyperplanes in Zariski topology that faces are also close. half-hyperplanes are not hyperplanes.

Let's think about your figure. Each of the line segments are not closed in Zariski Toplogy since the only closed subsets of a line in Zariski Topology except the whole line, are finite subsets. Hence saying that your $(n-2)$-skeleton is finite union of them doesn't prove it is closed.