Exercise 1.1 in Davey & Priestley, Introduction to Lattices and Order (1st edition) reads:

Let $P$ be a set on which a binary relation $<$ is defined such that, for all $x, y, z \in P,$

(a) $x < x$ is false,

(b) $x < y$ and $y < z$ imply $x < z.$

Prove that if $\leqslant$ is defined by $$x \leqslant y \iff (x < y \text{ or } x = y),$$ then $\leqslant$ is an order on $P,$ and moreover every order on $P$ arises from a relation $<$ on $P$ satisfying (a) and (b). [A binary relation satisfying (a) and (b) is called a strict order.]

A strict weak ordering on $P$ satisfies (a) and (b) together with:

(c) If $x < z$ then ($x < y$ or $y < z$).

I wish to define the associated total preorder on $P$ by: $$x \leqq y \iff\lnot(y < x),$$ and the associated equivalence relation on $P$ by: $$x \cong y \iff\lnot(x < y \text{ or } y < x) \iff (x \leqq y \text{ and } y \leqq x).$$

(This is because I prefer to avoid the ugly use of "wavy" symbols such as $\lesssim$.)

But might my proposed use of notation be confusing, even though it avoids the confusion that would certainly result - at least if Davey & Priestley's book is anything to go by - from the use of $\leqslant$ or $\leq$?

  • $\begingroup$ Yes it is acceptable since you defined them. One could also use << for a relation that you define, even though it has been used for much smaller than. $\endgroup$ – William Elliot Jun 4 '17 at 1:00

I would not recommend $\leqq$ for this. The symbol very strongly suggests "$<$ or $=$", and it feels pretty strange to me to use it for anything else. So while you are of course allowed to use whatever notation you want as long as you define it clearly, I don't think this is a good choice if you want to avoid confusion. Your proposed use of $\cong$ looks fine though.


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