# Smallest amount of planes to enclose a closed space in extended projective geometry $\mathbb R^3_{\pm\infty}$

The smallest amount of planes to enclose a polyhedron is 4 in the euclidean $\mathbb R^3$ where it encloses a tetrahedron. What is the smallest amount of planes to enclose a closed space in extended euclidean geometry aka projective geometry where you suppose infinities as points?

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By "plane" you mean an affine subspace with codimension 1, that is, a line in 2D or a hyperplane in >3 dimensions, right? – Henning Makholm Feb 11 '13 at 20:43
@HenningMakholm good question. After some googling -- yes, it must be affine but not yet sure about the codimension thing -- have to read more about it to be able to answer. – hhh Feb 11 '13 at 20:49
Once you fix a plane at infinity, projective geometry contains Euclidean geometry as a subset, so you can still form that same tetrahedron in projective space. On the other hand, unless you fix that plane at infinity, concepts like “inside”, “closed”, “compact”, “halfspace” and similar are meaningless, so you might have to rephrase your question and make explicit what definition you have in mind. – MvG Feb 13 '13 at 16:21
@MvG yes but what is the smallest amount of planes? I have understood that the the line is a large circle in projective geometry so spanning an area. So a plane would be a torus? So only one plane required in projective geometry where you consider infinities as points, right? – hhh Feb 13 '13 at 18:42