Ceiling defines Equivalence Relation

Theorem
Let $\mathcal R$ be the relation defined on $\R$ such that:
 * $\forall x, y, \in \R: \left({x, y}\right) \in \mathcal R \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$

where $\left \lceil {x}\right \rceil$ is the ceiling of $x$.

Then $\mathcal R$ is an equivalence, and $\forall n \in \Z$, the $\mathcal R$-class of $n$ is the half-open interval $\left({n - 1 \,.\,.\, n}\right]$.

Proof
Checking in turn each of the critera for equivalence:

Reflexivity

 * $\forall x \in \R: \left \lceil {x}\right \rceil = \left \lceil {x}\right \rceil$

Thus the ceiling function is reflexive.

Symmetry

 * $\forall x, y \in \R: \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil \implies \left \lceil {y}\right \rceil = \left \lceil {x}\right \rceil$

Thus the ceiling function is symmetric.

Transitivity
Let:
 * $\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$
 * $\left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$

Let:
 * $n = \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$

which follows from transitivity of $=$.

Thus:
 * $x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \left[{0 \,.\,.\, 1}\right)$

from Real Number is Ceiling minus Difference‎.

So:
 * $x = n - t_x$

and:
 * $z = n - t_z$

and so:
 * $\left \lceil {x}\right \rceil = \left \lceil {z}\right \rceil$

Thus the ceiling function is transitive.

Thus we have shown that $\mathcal R$ is an equivalence.

Now we show that the $\mathcal R$-class of $n$ is the interval $\left({n - 1 \,.\,.\, n}\right]$.

Defining $\mathcal R$ as above, with $n \in \Z$: