I want to understand order relations using their underlying implication mechanics and what this means for certain results, specifically looking at preference relations.

Using the logical rules of implication Is the following correct:

  • $x \ge y \implies x > y \lor x = y$?

Or should this be:

  • $x \ge y \iff x > y \lor x = y$

Because surely we also have:

  • $x > y \implies x \ge y$
  • $x = y \implies x \ge y$

Is it appropriate to think of order relations in this way?


In the context of preference relations. Does the logical i laid out for the general order relations above apply exactly to preferences If i were to change the following:

  • $\ge$ for $\succeq$
  • $>$ for $\succ$
  • $=$ for $\sim$

For example: An exam question notes that:

By definition $x \succ y$ means that $x \succeq y$

Intuitively this makes sense, if the preference is strict, then i can see that means it must also satisfy the weaker condition. Can I write this using implications as before? i.e.

$x \succ y$ is sufficient for $x \succeq y$ i.e. $x \succ y \implies x \succeq y$

Specific Questions:

  1. $x \succeq y \iff x \succ y \lor x \sim y$ ?

  2. $x \sim y \implies x \succeq y$ ?

  3. The same exam commentary gives the proof that we cannot have $x \succ x$ because:

$x \succeq y \; \forall x,y$ therefore there is no case where $x \succ x$

  • But surely if $x \succeq x \iff x \succ x \lor x \sim x$
  • Then using their own language $x \succ x$ means that $x \succeq x$.
  • I.e. as alluded to above we'd have $x \succ x \implies x \succeq x$
  • So why is their proof valid? The statement $x \succeq x$ doesn't seem to invalidate or contradict $x \succ x$?
  • 1
    $\begingroup$ Are you sure your citations are complete? By definition $x \succ y$ means that $x \succeq y$ is either false (if "means that" is read as "iff") or true (if "means that" is read as "implies"). But $x \succeq y \; \forall x,y$ makes no sense at all, since it only holds for the trivial preference relation where all alternatives are indifferent. $\endgroup$
    – VARulle
    Commented May 12, 2023 at 8:21

2 Answers 2


Just a sketch of a more general context, regarding relations in set theory.$^1$

You wrote:

I want to understand order relations [...] and what this means for certain results, specifically looking at preference relations.

[...]Does the logical I laid out for the general order relations above apply exactly to preferences If I were to change the following:

  • $\ge$ for $\succeq$
  • $>$ for $\succ$
  • $=$ for $\sim$

If you mean that any statement that holds for general order relations can be extended to preference relations, changing $\ge$ for $\succeq$, $>$ for $\succ$ and $=$ for $\sim$, no, this is not possible in general, this makes no sense.

Because we mustn't make confusion between order relations and preferences (or preference relations), they may have different properties that cannot be transferred to each other.

They are not the same thing: a preference relation isn't an order relations as defined in mathematics, it is a so-called preorder.

Relations, order relations, preorder, equivalence relations ( $\sim$ is an equivalence relation) are concepts belonging to set theory, together with properties as reflexivity, symmetry, anti-symmetry, transitivity and so on that can define the various concepts.

A preference is defined usually as a complete preorder so that you cannot apply to it automatically the properties of order relations.

In short, you must be careful, and, if necessary, refer to the rigorous definitions.


Just to give you an idea:

Definition Let $P$ be a set. An order (or partial order) on $P$ is a binary relation $\geq$ on $P$ such that, for all $x$, $y$, $z$ $\in P$,

(i) $x\geq$ x,

(ii) $x\geq y$ and $y \geq x$ implies $x=y$,

(iii) $x\geq y$ and $y\geq z $ implies $x \geq z$.

These conditions are referred to, respectively, as reflexivity, antisymmetry and transitivity.

An order relation is said total if, for all $x,y \in P$, or $x\geq y$ or $y \geq x$, that is any pair of elements of the set $P$ is comparable.

A relation which is reflexive and transitive but not necessarily antisymmetric is called a pre-order (or quasi-order).

Preferences are defined as a complete preorder and they can lack (and usually do) of antisymmetry.$^2$

Completeness for preference relations, in a set $Q$, is the same as total for order relations: for all $x,y \in Q$, or $x\succeq y$ or $y \succeq x$, that is any pair of elements is comparable.


An order relation $\geq$ on $P$ gives rise to a relation $>$ of strict inequality (strict order) : $x > y$ if and only if $x\geq y$ and $x\neq y$.

A similar induced relation, $\succ$, is defined for a preference relation, as explained in the answer by Herr K.

$^1$ In set theory, a relation on a set $A$ is defined as a subset $B$ of the cartesian product $A\times A$.

$^2$ If you think of the theory of the consumer, you can guess how antisymmetry can be disruptive for that theory as we know it.

  • 1
    $\begingroup$ Thanks again for this one, i agree that one of my questions makes no sense. Otherwise the difference in notation would be totally redundant! $\endgroup$
    – CormJack
    Commented May 20, 2023 at 21:42
  • 1
    $\begingroup$ I studied in reasonable detail order and partial orders last year, maybe i have just forgot my notes, I'll review and let you know if i still have any burning questions! $\endgroup$
    – CormJack
    Commented May 20, 2023 at 21:46

The definition of $x\succ y$ is: "$x\succsim y$ and not the case that $y\succsim x$". Formally, $$ x\succ y \quad\Longleftrightarrow\quad x\succsim y \;\wedge\; \lnot(y\succsim x) $$ By this definition, the exam question you quoted seems to have omitted the second half of a conjunct even though it says "by definition".

Also by the above definition, it's easy to see that the statement $$ x\succ x \quad\Longleftrightarrow\quad x\succsim x \;\wedge\; \lnot(x\succsim x) $$ can never be true, since the RHS of the bi-implication is always false.

  • $\begingroup$ Thank you very much for this! It was helpful, i realise they had included this second part, but due to my lack of understanding i had omitted it. I now see the importance! Thank you! $\endgroup$
    – CormJack
    Commented May 20, 2023 at 21:45

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