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I was doing this exercise:

Let $\sim$ be an indifference relation on X. For all $x \in X$ define $I(x) = \{y \in X: x\sim y\}$ Show that the set (of sets) $\{I(x): x \in X\}$ is a partition of $X$, i.e.

i) for all $x,y \in X$ either $I(x) = I(y)$ or $I(x) \cap I(y) = \emptyset$

ii) for every $x \in X$, there is $y \in X$ such that $x \in I(y)$


So, correct me if I'm wrong,
$I(x) = \{y \in X: x\sim y\}$ is the set of all the images of $x$, i.e. it contains all the $y$ such that they have an indifference relation with $x$ and $I(x) \subseteq X$

$\{I(x): x \in X\}$ is the collection of all different I-equivalence classes

then my answers to the two points are:

i) Since equivalence classes are disjoint we can either have:
(1) they belong to the same equivalence class, so it doesn't matter which element we choose as representative for that class, i.e. $I(x) = I(y)$ for all $x,y \in X$
(2) they belong to different equivalence classes so they belong to disjoint sets and $I(x) \cap I(y) = \emptyset$

ii) By completeness either $x \sim y$ or $y \sim x$

Are my answers wrong? How can I write them "formally"?

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    $\begingroup$ You start with what you call "a preference on $X$". What is that by definition? Secondly in the rest this preference (whatever it is) is not mentioned anymore. Then why was it mentioned? Thirdly: what exactly is an "indifference relation"? $\endgroup$ – drhab Sep 12 '18 at 10:18
  • $\begingroup$ It's an error in the text of the exercise, I just copied it from a book so I didn't even notice. I'll delete it. $\endgroup$ – Zhang_anlan Sep 12 '18 at 10:25
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    $\begingroup$ Is it so that $\sim$ denotes an equivalence relation on $X$? If so then that should be mentioned in your question. $\endgroup$ – drhab Sep 12 '18 at 10:26
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    $\begingroup$ Is "indifference relation" the same thing as equivalence relation? $\endgroup$ – drhab Sep 12 '18 at 10:31
  • $\begingroup$ Yes, the definition is: $x \sim y$ if and only if $x \geq y$ and $y \geq x$ $\endgroup$ – Zhang_anlan Sep 12 '18 at 10:45
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i)

If $z\in I(x)\cap I(y)$ then $x\sim z\wedge y\sim z$.

The relation $\sim$ is symmetric so that $x\sim z\wedge z\sim y$.

The relation $\sim$ is transitive so that $x\sim y$.

Now if $u\in I(y)$ or equivalently $y\sim u$ it follows again by transitivity that also $x\sim u$ or equivalently $u\in I(x)$.

Proved is now that $I(y)\subseteq I(x)$ and similarly it can be proved that $I(x)\subseteq I(y)$.

ii)

$y=x$ serves well here since $x\in I(x)$. This because the relation $\sim$ is reflexive.

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