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I've met a problem in M.A.Armstrong's Basic Topology.

If $K$ and $L$ are complexes in $\mathbb{E}^n$, show that $\vert K\vert\cap\vert L\vert$ is a polyhedron.

where $\vert K\vert$ and $\vert L\vert$ are the polyhedron of $K$ and $L$.

I think it's not hard to imagine this statement. But I can't find a formal proof for this.

Can you please help? Thank you.

EDIT: The definition of a polyhedron in Basic Topology is:

... the union of the simplexes which make up a particular complex is a subset of a euclidean space, and can therefore be made into a topological space by giving it the subspace topology. A complex $K$, when regarded in this way as a topological space, is called a polyhedron and written $\vert K\vert$.

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What if the two polyhedra touch each other's faces? –  J. M. May 10 '11 at 2:58
    
What does |X| is the polyhedron of X mean? –  M.B. May 10 '11 at 3:38
    
@M.B. : I've edited the post for explanation. –  Roun May 10 '11 at 4:40

1 Answer 1

Based upon very little knowledge, here is my guess: If a polyhedron is the intersection of half-spaces, then the intersection of two polyhedrons would be the intersection of all the half-spaces in both, and would thus be a polyhedron.

This seems too easy, so I am sure I have stepped into something, but I don't know what.

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Except for "polyhedrons" (I know it is said to be correct) your answer is perfectly fine :) –  t.b. Aug 1 '11 at 5:20
    
@marty:Thank you for your idea. Well, the definition of a polyhedron given in the problem may be different from yours. Here it is not necessarily intersection of half-spaces, since it may not be convex. –  Roun Aug 2 '11 at 11:01

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