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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 $B$.

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For $z\in\mathbb{Z}$, let $$A_z=\{(x,y)\in\mathbb{R}^2: x>z-1\mbox{ and }y<z+1\}.$$ 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.)

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  • $\begingroup$ In other words, it's a countable union of bottom-left quadrants. $\endgroup$ Jul 7 at 22:28
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    $\begingroup$ @GregMartin *bottom-right $\endgroup$
    – Milten
    Jul 7 at 22:45
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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).

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  • $\begingroup$ Hi, I took the liberty of fixing up your sets. $\endgroup$
    – Milten
    Jul 7 at 23:13

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