Parallelism is Equivalence Relation

Theorem

Let $S$ be the set of straight lines in the plane.

For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is parallel to $l_2$.


Then $\parallel$ is an equivalence relation on $S$.


Proof

Checking in turn each of the criteria for equivalence:


Reflexivity

By definition of parallel lines, the contemporary definition is for a straight line to be declared parallel to itself.

Hence for a straight line $l$:

$l \parallel l$

Thus $\parallel$ is seen to be reflexive.

$\Box$


Symmetry

Let $l_1 \parallel l_2$.

By definition of parallel lines, $l_1$ does not meet $l_2$ when produced indefinitely.

Hence $l_2$ similarly does not meet $l_1$ when produced indefinitely.

That is:

$l_2 \parallel l_1$

Thus $\parallel$ is seen to be symmetric.

$\Box$


Transitivity

From Parallelism is Transitive Relation:

$l_1 \parallel l_2$ and $l_2 \parallel l_3$ implies $l_1 \parallel l_3$.

Thus $\parallel$ is seen to be transitive.

$\Box$


$\parallel$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$


Sources

  • 1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $1$. Elementary Operations on Sets
  • 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.3$: Equivalence Relations: $(3)$
  • 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): equivalence relation
  • 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): equivalence relation