Hilbert’s third problem and what a polyhedron is [closed]

Question

What is the definition of a polyhedron used by Hilbert’s third problem?

Answer 1

The question itself is interesting, because in historical contexts (sometimes even modern ones) it is not always clear what a polyhedron means. To anyone who thinks this is evident from the outset, I would encourage having a look at Lakatos's entertaining 1976 book Proofs and Refutations. Before reading the book, I thought I had a pretty clear idea of the possible problem cases where Euler's formula $V-E+F=2$ fails, but I had myself surprised.

In Hilbert's case we should look into the original text. His question starts with tetrahedra. To wit, original text, as printed in Göttinger Nachrichten, 1900, pp. 253–297,

Ein solcher wäre erbracht, sobald es gelingt, zwei Tetraeder mit gleicher Grundfläche und von gleicher Höhe anzugeben, die sich auf keine Weise in congruente Tetraeder zerlegen lassen und die sich auch durch Hinzufügung congruenter Tetraeder nicht zu solchen Polyedern ergänzen lassen, für die ihrerseits eine Zerlegung in congruente Tetraeder möglich ist.

and English translation by Mary Newson,

This would be obtained, as soon as we succeeded in specifying two tetrahedra of equal bases and equal altitudes which can in no way be split up into congruent tetrahedra, and which cannot be combined with congruent tetrahedra to form two polyhedra which themselves could be split up into congruent tetrahedra.

There are some points worth pondering here — which I did not notice at first!

1. Hilbert talks not only about splitting the tetrahedra, but also about combining them into polyhedra. From the context, it seems a polyhedron here is anything you can obtain by such combination; for example, holes are allowed and convexity is not required.

2. The wording with "splitting" and "combining" seems to imply that the interiors of pieces are disjoint. We are splitting and combining solids. (No star polyhedra!)

3. Although the word "finite" is conspicuously absent, it seems obviously implied that both splitting and combining is with finite number of pieces, because the whole point is to avoid "exhaustion".