In this discussion, the reader should be familiar with the ordinary Two-Piece Property (TPP). A quick overview of the TPP in the plane can be found here. We first define the GTPP in two dimentions:

Definition. A grid square is a subset S of R² with center (x₀, y₀) and radius r defined by

S = {(x, y): max(|x - x₀|, |y - y₀|) = r}.

We denote the closure of the interior of S by D₁ and the closure of the exterior by D₂. Note that a grid square is a circle in the maximum-value metric.

Definition. A subset A of R² has the Grid-Square Two-Piece Property (GTPP) if for every grid square S the intersection of A with D_i is path connected, i = 1, 2.

Clearly R² has the GTPP, since the interior and exterior of grid squares are connected. Also, the empty set, singleton sets, and sets consisting of exactly two points have the GTPP. (Note that these sets also have the ordinary TPP and the STPP.) We first have the following useful lemma:

Lemma. Let U be a bounded subset of R². Then the smallest grid rectangle (a rectangle whose edges are parallel to the axes) containing U is contained in the closure of the interior of every grid square that contains U.

proof. Suppose there is a grid square S with center (h,k) and radius r such that the closure of the interior of S contains U. By construction, all four edges of the rectangle contain points of U. Consider a vertex of the rectangle with coordinates (a,d). Then there are two points of U with coordinates (a,b) and (c,d) that lie on adjacent edges of the rectangle. Both points are in U and are thus in S, so the following hold:

max(|a - h|, |b - k|) <= r and
max(|c - h|, |d - k|) <= r.

Thus each |a - h|, |b - k|, |c - h|, |d - k| is less than or equal to r. We now see that the vertex itself is in S, since

max(|a - h|, |d - k|) <= r.

This argument may be repeated for each vertex, yielding the same result. Since S is convex, the convex hull of the vertices is also contained in S, which implies that the rectangle and its interior are in S.

Proposition 1. Every grid square in R² has the GTPP.

proof. First note that the GTPP is invariant under translations, scaling, and 90°-rotations, so it is sufficient to show that the unit grid square centered at the origin, which we will call A, has the GTPP. Since A is a simple closed curve, given any grid square S, if the intersection of A with D₁ is connected, then so is the intersection of A with D₂ (recall that D₁ is the closure of the interior of S and D₂ is the closure of the exterior of S). Thus we may consider only the interior intersection of A and D₁. There are five cases:

1. The interior intersection is empty.
2. The interior intersection is non-empty, but D₁ contains no vertices of A.
3. D₁ contains exactly one vertex of A.
4. D₁ contains exactly two vertices of A.
5. D₁ contains all four vertices of A, and so the interior intersection is all of A.

Note that these cases cover all possibilities, since if D₁ contained three vertices of A it would also contain the fourth. In cases 1 and 5, connectedness follows immediately.

In case 2, the interior intersection must be a horizontal or vertical segment (or a single point in the degenerate case). If the interior intersection contained points from adjacent edges of A, then the common vertex would be included (by the above Lemma), a contradiction. If the interior intersection contained points from opposite edges, then the radius of S must be at least 1. This implies that D₁ contains a point of one of the adjacent edges, and so the interior intersection must contain two vertices of A, again a contradiction.

In case 3, the interior intersection must be two segments, each from adjacent edges, joined at the single vertex that is contained in the intersection. The reasoning for this is similar to that of case 2.

In case 4, the interior intersection must be three segments, joined at the two vertices that are contained in the intersection. Again we can use similar reasoning to cases 2 and 3.

In every case the interior intersection is connected, so A has the GTPP.

Here is a demo that illustrates Propositions 1 and 2.

It is true in general that Grid Cubes have the GTPP. Note that rectangles do not have the GTPP, nor do non-grid squares. It is natural to ask whether there are any closed plane curves that have the GTPP other than grid squares themselves. The following proposition answers that question.

Proposition 2. Every closed curve in R² with the GTPP is a grid square.

proof. Let C be a closed curve in R², and let D be its bounding rectangle (the smallest rectangle whose interior's closure contains C). If D is not a square, then there is a grid square (shown below in red) that divides C into at least four connected components.

We will therefore assume that D is a square, and suppose that C has the GTPP. Suppose D has side length s. We may choose a grid square of side length s - d, for some small d > 0, concentric with D (shown below in red in (a)). Since C has the GTPP and we can make d as small as we want, it must in fact look like in (b) below, with the four points x, y, z, and w connected by a path that lies on D (the path is shown with C not on D between y and z, but it may be any of the four orientations).

Now we choose a grid square with side length s and translated as shown in (c) by a small distance d > 0. If C is as shown, it is broken into four pieces by the grid square, and therefore C must be as in (d); that is, C must be a grid square itself.

We now give a brief treatment of embedded curves and the GTPP.