Proposition 1'. Every grid cube in Rⁿ has the GTPP.

proof. We will consider, without loss of generality, the grid cube in Rⁿ centered at the origin with radius 1; call it A. Let S be an arbitrary grid cube, centered at x with radius r. Let D₁ be the closure of the interior of S, and D₂ be the closure of the exterior of S, as usual. We will first show that the intersection of A with D₁ (which we will call the interior intersection) is path-connected.

Consider two points a = (a₀, ..., a_n-1) and b = (b₀, ..., b_n-1) in the interior intersection. Since a and b are in A, their components satisfy |a_i - x_i| <= r and |b_i - x_i| <= r for every i = 0, ..., n-1. Additionally, there exist l, m such that |a_l| = |b_m| = 1. We consider the following three cases:

1. The points a and b lie on the same (n-1)-hypercube of A: l = m and a_l = b_m.
2. The points a and b lie on adjacent (n-1)-hypercubes of A: l != m.
3. The points a and b lie on opposite (n-1)-hypercubes of A: l = m and a_l = -b_m.

In each case we can show the existence of a path in the interior intersection connecting a and b:

1. Let p = (p₀, ..., p_n-1) satisfy min(a_i, b_i) <= p_i <= max(a_i, b_i) for every i. Then p is in the interior intersection. Every point on the segment joining a and b is of this form, so the segment itself is a valid path.

2. We must find a point c that lies on both the (n-1)-hypercube of A containing a and the one containing b. Once we have c, the path consisting of the segment from a to c and the segment from c to b will do the job. Take c to be (a₀, ..., a_m-1, b_m, a_m+1, ..., a_n-1), and we see that c has the desired properties.

3. Again, we can reduce this case to a previous one by choosing an intermediate point. Note that any point on one of the 2n-2 adjacent (n-1)-hypercubes will work; in particular, it is enough to find a point on one of the "hyperedges" of A. Let p_k+ = (a₀, ..., a_k-1, 1, a_k+1, ..., a_n-1) and p_k- = (a₀, ..., a_k-1, -1, a_k+1, ..., a_n-1). If k != l then either p_k+ or p_k- will work. We now show that at least one of these points must be in the interior intersection. Since the interior of S contains points on opposite (n-1)-hypercubes of A, we know that r >= 1. We now have three colinear points, p_k-, a, and p_k+, the middle of which is in D₁. However, the endpoints p_k- and p_k+ are exactly 2 units apart and D₁ is at least 2 units wide. Thus at least one of the endpoints must be included.

We have now shown that the interior intersection of A with D₁ is connected.

To prove the proposition it is now sufficient to show that the exterior intersection (the intersection of A with D₂) is connected. We proceed by induction on n: we have already shown the case of n = 2. We may assume that the intersection of S with the boundary of any hyperface of A is connected, since this is the n-1 case; if follows that the intersection of S with the hyperface itself is connected. Now we show that the entire exterior intersection (the union of these subintersections) consists of one connected component.

We first see that if the exterior intersection is nonempty, then at least one of the hyperfaces of A must lie entirely within the exterior intersection: choose a point p = (p₀, p₁, ..., p_n-1) in the exterior intersection. For some i, |p_i - x_i| >= r, but since p is on A, we also have |p_i| <= 1. Thus, either |1 - x_i| >= r or |-1 - x_i| >= r, and so every point on A with a 1 (or every point with a -1) in the ith component (i.e. an entire hyperface of A) must be in the exterior intersection.

Let A₁ be such a hyperface, and A₁' be the opposite hyperface. Since A₁ contains points of each adjacent hyperface, all points in the exterior intersection on all faces of A except A₁' form a connected set. We must now consider points outside of S on A₁'. If such points exist, then there must be such points on the boundary of A₁', which are shared with the adjacent faces. Now the connectivity of the points in the exterior intersection on A₁' shows the connectivity of all points in the exterior intersection, and the proof is complete.

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