How to Find the Intersection of Two Polyhedra in Topology

Hello! This is a beautiful problem in piecewise-linear topology. Let's work through understanding how to show that the intersection of two polyhedra remains a polyhedron.

What We're Solving:

We need to prove that if P and Q are polyhedra (in the sense of Rourke & Sanderson), then their intersection P ∩ Q is also a polyhedron. The hint suggests using ε-neighbourhoods, which is key to working with their definition of polyhedra.

The Approach:

In Rourke & Sanderson, a polyhedron is defined as a subset of Euclidean space that locally "looks like" the intersection of finitely many half-spaces. The ε-neighbourhood hint is crucial because their definition involves showing that around each point, there exists some ε > 0 such that the ε-neighbourhood of that point intersected with the polyhedron gives us this half-space intersection structure.

The strategy is to show that P ∩ Q satisfies this local condition at every point.

Step-by-Step Solution:

Step 1: Set up what we need to prove For any point x ∈ P ∩ Q, we need to show there exists ε > 0 such that in the ε-neighbourhood N_ε(x), the set (P ∩ Q) ∩ N_ε(x) can be expressed as the intersection of finitely many closed half-spaces with N_ε(x).

Step 2: Use the polyhedron property of P and Q Since x ∈ P and P is a polyhedron, there exists ε₁ > 0 such that P ∩ N_ε₁(x) is the intersection of finitely many half-spaces with N_ε₁(x). Let's call these half-spaces H₁, H₂, ..., H_m.

Similarly, since x ∈ Q and Q is a polyhedron, there exists ε₂ > 0 such that Q ∩ N_ε₂(x) is the intersection of finitely many half-spaces with N_ε₂(x). Let's call these half-spaces K₁, K₂, ..., K_n.

Step 3: Choose the right neighbourhood Let ε = min(ε₁, ε₂). This ensures that both the P and Q polyhedron properties hold in N_ε(x).

Step 4: Express the intersection Now, in N_ε(x), we have:

• P ∩ N_ε(x) = (H₁ ∩ ... ∩ H_m) ∩ N_ε(x)
• Q ∩ N_ε(x) = (K₁ ∩ ... ∩ K_n) ∩ N_ε(x)

Therefore: (P ∩ Q) ∩ N_ε(x) = (H₁ ∩ ... ∩ H_m ∩ K₁ ∩ ... ∩ K_n) ∩ N_ε(x)

Step 5: Conclude We've expressed (P ∩ Q) ∩ N_ε(x) as the intersection of finitely many half-spaces (the H's and K's) with N_ε(x). This is exactly what we needed to show that P ∩ Q satisfies the local polyhedron condition at x.

Since x was arbitrary in P ∩ Q, this holds for all points, so P ∩ Q is a polyhedron.

The Answer:

P ∩ Q is a polyhedron because locally around each point, it can be expressed as the intersection of finitely many half-spaces (those defining P plus those defining Q in appropriate neighbourhoods).

Memory Tip:

Think of it this way: "Local constraints add up!" If P is locally defined by some linear inequalities and Q is locally defined by other linear inequalities, then P ∩ Q is locally defined by all those inequalities together. The intersection doesn't create anything more complex than what we started with.

The key insight is that intersecting polyhedra can't make things "more curved" or "more complicated" - you're just adding more linear constraints, which keeps everything piecewise-linear!