# How can $\{(x,y)\in\mathbb{R}^2:\lceil x\rceil\geq\lfloor y\rfloor\}$ be analytically characterized, if at all?

I am interested in analytically characterizing $$\Omega = \{(x,y)\in\mathbb{R}^2:\lceil x\rceil\geq\lfloor y\rfloor\}.$$ By "analytically characterize" I actually mean "Expressing $$\Omega$$ as a set -- or a (preferably finite) union/intersection of sets -- defined without using floor, ceiling, or integer/fractional part functions." (Apologies for the abuse of terminology.)

Edit: Clearly $$A=\{(x,y)\in\mathbb{R}^2: x-y>-1\}$$ is a proper subset of $$\Omega$$. I think $$\Omega\setminus A$$ is some proper subset $$B\subset\{(x,y)\in\mathbb{R}^2: -1\leq x-y<2\}$$; the challenging part for me is characterizing $$B$$.

For $$z\in\mathbb{Z}$$, let $$A_z=\{(x,y)\in\mathbb{R}^2: x>z-1\mbox{ and }y Note that we have $$(x,y)\in A_z$$ iff $$\lceil x\rceil\ge z$$ and $$\lfloor y \rfloor\le z$$. Consequently, we have $$(x,y)\in\Omega\iff (x,y)\in A_{\lfloor y\rfloor}\iff (x,y)\in A_{\lceil x\rceil}.$$ Consequently we can express $$\Omega$$ as a countable union of sets defined via simple inequalities: $$\Omega=\bigcup_{z\in\mathbb{Z}} A_z.$$
Interestingly, we can also prove a negative result relating to this question - namely, that $$\Omega$$ is not a semialgebraic set. This takes some work, however. (More generally, $$\Omega$$ is not definable in any o-minimal structure on $$\mathbb{R}$$, but both "definable" and "o-minimal" are technically complicated notions.)
Note that $$\cos(\pi(x-y))= \cos(\pi(x+y))$$ holds when either $$\pi(x+y)-\pi(x-y)$$ or $$\pi(x+y)+\pi(x-y)$$ is a multiple of $$2\pi$$, i.e., when one of $$x,y$$ is $$\in\Bbb Z$$. One then quickly sees that $$\{\,(x,y)\in\Bbb R^2\mid \cos(\pi(x-y))\le \cos(\pi(x+y))\,\}$$ is a (closed) checkerboard pattern in the lattice grid. If you cut this with the diagonal halfplane $$\{\,(x,y)\mid x-y\ge -2\,\}$$ and take the union with another diagonal half plane $$\{\,(x,y)\mid x-y\ge-1\,\},$$ you get exactly the "zigzagged halfplane" we want. (You may need to adjust some translation perhaps - I didn't check).