<p>The <em>definition</em> of <span class="math-container">$x\succ y$</span> is: &quot;<em><span class="math-container">$x\succsim y$</span> and not the case that <span class="math-container">$y\succsim x$</span></em>&quot;. Formally,
<span class="math-container">$$
x\succ y \quad\Longleftrightarrow\quad x\succsim y \;\wedge\; \lnot(y\succsim x)
$$</span>
By this definition, the exam question you quoted seems to have omitted the second half of a conjunct even though it says &quot;by definition&quot;.</p>
<p>Also by the above definition, it's easy to see that the statement
<span class="math-container">$$
x\succ x \quad\Longleftrightarrow\quad x\succsim x \;\wedge\; \lnot(x\succsim x)
$$</span>
can never be true, since the RHS of the bi-implication is always false.</p>

CC BY-SA 4.0
2023-05-20T21:45:11.910Z
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!
CormJack
CC BY-SA 4.0
2023-05-11T18:59:22.217Z
2023-05-11T18:59:22.217Z
Herr K.
<p>Just a sketch of a more general context, regarding <em><strong>relations in set theory</strong></em>.<span class="math-container">$^1$</span></p>
<p>You wrote:</p>
<blockquote>
<p>I want to understand order relations [...] and
what this means for certain results,
specifically looking at preference relations.</p>
<p>[...]Does the logical I laid out
for the general order relations above apply exactly to preferences If
I were to change the following:</p>
<ul>
<li><span class="math-container">$\ge$</span> for <span class="math-container">$\succeq$</span></li>
<li><span class="math-container">$&gt;$</span> for <span class="math-container">$\succ$</span></li>
<li><span class="math-container">$=$</span> for <span class="math-container">$\sim$</span></li>
</ul>
</blockquote>
<p>If you mean that any statement that holds for general order relations can be extended  to preference relations, changing <span class="math-container">$\ge$</span> for <span class="math-container">$\succeq$</span>, <span class="math-container">$&gt;$</span> for <span class="math-container">$\succ$</span>
and  <span class="math-container">$=$</span> for <span class="math-container">$\sim$</span>, no, this is not possible in general, this makes no sense.</p>
<p>Because we mustn't make confusion between <em><strong>order relations</strong></em> and <em><strong>preferences</strong></em> (or <em><strong>preference relations</strong></em>), they may have different properties that cannot be transferred to each other.</p>
<p>They are not the same thing: <strong>a preference relation</strong>  isn't  an <em>order relations</em>  as defined in mathematics, <strong>it is a so-called <em>preorder</em>.</strong></p>
<p><em>Relations</em>, <em>order relations</em>,  <em>preorder,</em> <em>equivalence relations</em> ( <span class="math-container">$\sim$</span> is an equivalence relation)  are  concepts belonging to <strong>set theory</strong>, together with properties as <em>reflexivity, symmetry, anti-symmetry, transitivity</em> and so on that can define the various concepts.</p>
<p><strong>A preference is defined usually as a <em>complete preorder</em></strong> so that you cannot apply to it automatically the properties of order relations.</p>
<p>In short, you must be careful, and, if necessary, refer to the rigorous definitions.</p>
<p><span class="math-container">$\;$</span></p>
<p>Just to give you an idea:</p>
<p><em><strong>Definition</strong></em> Let <span class="math-container">$P$</span> be a set. An <em><strong>order</strong></em> (or <em><strong>partial order</strong></em>) on <span class="math-container">$P$</span> is a binary relation <span class="math-container">$\geq$</span> on <span class="math-container">$P$</span> such that, for all <span class="math-container">$x$</span>, <span class="math-container">$y$</span>, <span class="math-container">$z$</span> <span class="math-container">$\in P$</span>,</p>
<p>(i) <span class="math-container">$x\geq$</span> x,</p>
<p>(ii) <span class="math-container">$x\geq y$</span> and <span class="math-container">$y \geq x$</span> implies <span class="math-container">$x=y$</span>,</p>
<p>(iii) <span class="math-container">$x\geq y$</span> and <span class="math-container">$y\geq z $</span> implies <span class="math-container">$x \geq z$</span>.</p>
<p>These conditions are referred to, respectively, as <strong>reflexivity, antisymmetry</strong> and <strong>transitivity</strong>.</p>
<p>An order relation is said <em><strong>total</strong></em> if, for all <span class="math-container">$x,y \in P$</span>, or <span class="math-container">$x\geq y$</span> or <span class="math-container">$y \geq x$</span>, that is any pair of elements of the set <span class="math-container">$P$</span> is comparable.</p>
<p>A relation which is reflexive and transitive but not necessarily antisymmetric is called a <strong>pre-order</strong> (or <strong>quasi-order</strong>).</p>
<p>Preferences are defined as a <strong><em>complete preorder</em> and they can lack  (and usually do)  of antisymmetry.</strong><span class="math-container">$^2$</span></p>
<p><em>Completeness</em> for preference relations, in a set <span class="math-container">$Q$</span>,   is the same as <em>total</em> for order relations:  for all <span class="math-container">$x,y \in Q$</span>, or <span class="math-container">$x\succeq y$</span> or <span class="math-container">$y \succeq x$</span>, that is any pair of elements is comparable.</p>
<p><span class="math-container">$\;$</span></p>
<p>An order relation <span class="math-container">$\geq$</span> on <span class="math-container">$P$</span> gives rise to a relation <span class="math-container">$&gt;$</span> of strict inequality (<em>strict order</em>) : <span class="math-container">$x &gt; y$</span> if and only if <span class="math-container">$x\geq y$</span> and <span class="math-container">$x\neq y$</span>.</p>
<p><strong>A similar induced relation, <span class="math-container">$\succ$</span>, is defined for a preference relation</strong>, as explained in the answer by Herr K.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup> In set theory, a <em>relation</em> on a set <span class="math-container">$A$</span> is defined as a subset <span class="math-container">$B$</span> of the cartesian product <span class="math-container">$A\times A$</span>.</sup></p>
<p><span class="math-container">$^2$</span> <sup> If you think of the theory of the consumer, you can guess how antisymmetry can be disruptive for that theory as we know it.</sup></p>

CC BY-SA 4.0
2023-05-20T21:42:22.240Z
Thanks again for this one, i agree that one of my questions makes no sense. Otherwise the difference in notation would be totally redundant!
CormJack
CC BY-SA 4.0
2023-05-20T21:46:06.047Z
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!
CormJack
CC BY-SA 4.0
2023-05-12T21:34:53.210Z
2023-05-12T21:34:53.210Z
BakerStreet
<p>I want to understand order relations using their underlying implication mechanics and what this means for certain results, specifically looking at preference relations.</p>
<p>Using the logical rules of implication Is the following correct:</p>
<ul>
<li><span class="math-container">$x \ge y \implies x &gt; y \lor x = y$</span>?</li>
</ul>
<p>Or should this be:</p>
<ul>
<li><span class="math-container">$x \ge y \iff x &gt; y \lor x = y$</span></li>
</ul>
<p>Because surely we also have:</p>
<ul>
<li><span class="math-container">$x &gt; y \implies x \ge y$</span></li>
<li><span class="math-container">$x = y \implies x \ge y$</span></li>
</ul>
<p>Is it appropriate to think of order relations in this way?</p>
<p><strong>Preferences:</strong></p>
<p>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:</p>
<ul>
<li><span class="math-container">$\ge$</span> for <span class="math-container">$\succeq$</span></li>
<li><span class="math-container">$&gt;$</span> for <span class="math-container">$\succ$</span></li>
<li><span class="math-container">$=$</span> for <span class="math-container">$\sim$</span></li>
</ul>
<p>For example:
An exam question notes that:</p>
<blockquote>
<p><em>By definition <span class="math-container">$x \succ y$</span> means that <span class="math-container">$x \succeq y$</span></em></p>
</blockquote>
<p>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.</p>
<p><span class="math-container">$x \succ y$</span> is sufficient for <span class="math-container">$x \succeq y$</span>
i.e. <span class="math-container">$x \succ y \implies x \succeq y$</span></p>
<p><strong>Specific Questions:</strong></p>
<ol>
<li><p><span class="math-container">$x \succeq y \iff x \succ y \lor x \sim y$</span> ?</p>
</li>
<li><p><span class="math-container">$x \sim y \implies x \succeq y$</span>  ?</p>
</li>
<li><p>The same exam commentary gives the proof that we cannot have <span class="math-container">$x \succ x$</span> because:</p>
</li>
</ol>
<blockquote>
<p><span class="math-container">$x \succeq y \; \forall x,y$</span> therefore there is no case where  <span class="math-container">$x \succ x$</span></p>
</blockquote>
<ul>
<li>But surely if <span class="math-container">$x \succeq x \iff x \succ x \lor x \sim x$</span></li>
<li>Then using their own language <span class="math-container">$x \succ x$</span> <em>means that</em>  <span class="math-container">$x \succeq x$</span>.</li>
<li>I.e. as alluded to above we'd have <span class="math-container">$x \succ x \implies x \succeq x$</span></li>
<li>So why is their proof valid? The statement <span class="math-container">$x \succeq x$</span> doesn't seem to invalidate or contradict <span class="math-container">$x \succ x$</span>?</li>
</ul>

CC BY-SA 4.0
2023-05-12T08:21:53.940Z
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.
VARulle
<microeconomics><mathematical-economics><preferences><proof>
Order relations and preferences using logic
CormJack
# Order relations and preferences using logic

(Asked by: username_0 on 2023-05-11)

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?

**Preferences:**

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 $x \succ x$?

**Tags:** microeconomics, mathematical-economics, preferences, proof
> **Comments:**
> **username_1**: 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.  

---

## 2 answers

### Answer by username_3 on 2023-05-12. Score: 5 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks again for this one, i agree that one of my questions makes no sense. Otherwise the difference in notation would be totally redundant!  
> **username_0**: 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!  

---

### Answer by username_2 on 2023-05-11. Score: 5

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.

> **Comments:**
> **username_0**: 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!  

---

