<p>The usual definition of <span class="math-container">$x \succeq y$</span> is that either <span class="math-container">$x \sim y$</span> or <span class="math-container">$x \succ y$</span> hold. Your proof and the assumption of completeness are not necessary.</p>
<hr />
<p>One could also start from <span class="math-container">$\succeq$</span> and <span class="math-container">$\sim$</span> and define <span class="math-container">$x \succ y$</span> in such a way that <span class="math-container">$x \succeq y$</span> holds but <span class="math-container">$x \sim y$</span> does not. Even in this case, it is true without any additional assumptions that <span class="math-container">$x \succeq y$</span> implies either <span class="math-container">$x \sim y$</span> or <span class="math-container">$x \succ y$</span>.</p>

CC BY-SA 4.0
2022-03-18T09:58:19.207Z
@VARulle I guess "usual" depends on what you read, but you are probably right. Also, did you happen to see the part of my answer under the line horizontal line?
Giskard
CC BY-SA 4.0
2022-03-18T10:44:34.900Z
I happened to see that lower part, but I chose to comment on the upper part, since the "usual definition" of $x \succeq y$ is quite standard. "Starting from $\succeq$ and $\sim$", as the lower part suggests, is only possible after already having defined $\sim$, of course.
VARulle
CC BY-SA 4.0
2022-03-18T09:01:42.570Z
Late comment: The usual definition of $x \succeq y$ is not that either $x \sim y$ or $x \succ y$ hold. The relation $\succeq$ is what is given, and then $x \sim y$ and $x \succ y$ are defined via ($x \succeq y$ and $y \succeq x$) and ($x \succeq y$ and not($y \succeq x$)), respectively. So a proof, though easy, is necessary.
VARulle
CC BY-SA 4.0
2021-12-10T06:33:57.930Z
2021-12-10T06:33:57.930Z
Giskard
<p>Mas Collel Micro Theory question:</p>
<p>Suppose that X is a set. Let ≽ be a binary preference on X. And ~ represents indifference defined from ≽.</p>
<p>If ≽ satisfy completeness, is it okay to assume that:</p>
<p>x ≽ y implies either x &gt;y or x ~ y ?</p>
<p>My homemade proof:</p>
<p>≽ is complete -&gt; x ≽ y or y ≽ x</p>
<p>This implies either:</p>
<ol>
<li>x ≽ y &amp; not(y ≽ x) =&gt; x &gt; y</li>
<li>x ≽ y &amp; y ≽ x =&gt; x ~ y</li>
</ol>

<preferences><theory><proof>
Does x ≽ y imply x > y or x ~ y in preferences?
Wizard74
# Does x ≽ y imply x > y or x ~ y in preferences?

(Asked by: username_0 on 2021-12-09)

Mas Collel Micro Theory question:

Suppose that X is a set. Let ≽ be a binary preference on X. And ~ represents indifference defined from ≽.

If ≽ satisfy completeness, is it okay to assume that:

x ≽ y implies either x >y or x ~ y ?

My homemade proof:

≽ is complete -> x ≽ y or y ≽ x

This implies either:

1. x ≽ y & not(y ≽ x) => x > y
2. x ≽ y & y ≽ x => x ~ y

**Tags:** preferences, theory, proof

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## 1 answers

### Answer by username_1 on 2021-12-10. Score: 3 | Accepted answer

The usual definition of $x \succeq y$ is that either $x \sim y$ or $x \succ y$ hold. Your proof and the assumption of completeness are not necessary.

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One could also start from $\succeq$ and $\sim$ and define $x \succ y$ in such a way that $x \succeq y$ holds but $x \sim y$ does not. Even in this case, it is true without any additional assumptions that $x \succeq y$ implies either $x \sim y$ or $x \succ y$.

> **Comments:**
> **username_2**: Late comment: The usual definition of $x \succeq y$ is not that either $x \sim y$ or $x \succ y$ hold. The relation $\succeq$ is what is given, and then $x \sim y$ and $x \succ y$ are defined via ($x \succeq y$ and $y \succeq x$) and ($x \succeq y$ and not($y \succeq x$)), respectively. So a proof, though easy, is necessary.  
> **username_1**: @VARulle I guess "usual" depends on what you read, but you are probably right. Also, did you happen to see the part of my answer under the line horizontal line?  
> **username_2**: I happened to see that lower part, but I chose to comment on the upper part, since the "usual definition" of $x \succeq y$ is quite standard. "Starting from $\succeq$ and $\sim$", as the lower part suggests, is only possible after already having defined $\sim$, of course.  

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