Positive Difference Relation on Reals is Transitive

Theorem

Let $P \subseteq \R$ be a subset of the real numbers such that:

$(1): \quad 1 \in P$

$(2): \quad a, b \in P \implies a + b \in P$

$(3): \quad$ For all $x \in \R$, exactly one of these is true:

$x \in P$

$x = 0$

$-x \in P$

Let $Q \subseteq \R \times \R$ be the relation on $\R$ defined as:

$Q = \set {\tuple {a, b} \in \R: a - b \in P}$

Then $Q$ is a transitive relation.

Proof

Let $a - b \in P$ and $b - c \in P$.

By condition $(2)$:

$\paren {a - b} + \paren {b - c} \in P$

Simplifying:

$a - c \in P$

The result follows by definition of transitive relation.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $1$