Ceiling defines Equivalence Relation

Theorem

Let $\RR$ be the relation defined on $\R$ such that:

$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \ceiling x = \ceiling y$

where $\ceiling x$ is the ceiling of $x$.

Then $\RR$ is an equivalence, and $\forall n \in \Z$, the $\RR$-class of $n$ is the half-open interval $\hointl {n - 1} n$.

Proof

Checking in turn each of the criteria for equivalence:

Reflexivity

$\forall x \in \R: \ceiling x = \ceiling x$

Thus the ceiling function is reflexive.

$\Box$

Symmetry

$\forall x, y \in \R: \ceiling x = \ceiling y \implies \ceiling y = \ceiling x$

Thus the ceiling function is symmetric.

$\Box$

Transitivity

Let:

$\ceiling x = \ceiling y$

$\ceiling y = \ceiling z$

Let:

$n = \ceiling x = \ceiling y = \ceiling z$

which follows from transitivity of $=$.

Thus:

$x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \hointr 0 1$

from Real Number is Ceiling minus Difference‎.

So:

$x = n - t_x$

and:

$z = n - t_z$

and so:

$\ceiling x = \ceiling z$

Thus the ceiling function is transitive.

$\Box$

Thus we have shown that $\RR$ is an equivalence.

Now we show that the $\RR$-class of $n$ is the interval $\hointl {n - 1} n$.

Defining $\RR$ as above, with $n \in \Z$:

\(\ds x\)

\(\in\)

\(\ds \eqclass n \RR\)

\(\ds \leadsto \ \ \)

\(\ds \ceiling x\)

\(=\)

\(\ds \ceiling n = n\)

\(\ds \leadsto \ \ \)

\(\ds \exists t \in \hointr 0 1: \, \)

\(\ds x\)

\(=\)

\(\ds n - t\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds \hointl {n - 1} n\)

$\blacksquare$

Also see

• Floor defines Equivalence Relation