1. ## Equivalence Relation

Let $R$ be an equivalence relation on $P(X)$ by $ARB$ if $A\cup Y=B \cup Y$, where $X=\{1,2,3,4\}$, and $Y=\{3,4\}$.
What is the equivalence class of $\{1,2\}$?

I know there is only one element of the power set $X$ for the answer such that

$A \cup \{3,4\} = \{1,2\} \cup \{3,4\}=\{1,2,3,4\}$

For $A$, the possibilities are

$\{1,2\}$
$\{1,2,3\}$
$\{1,2,3,4\}$

Question: Since the intersection of any two equivalence classes is an empty set, does it mean that we get to pick only one of three possibilities listed above?

2. Originally Posted by novice

Question: Since the intersection of any two equivalence classes is an empty set, does it mean that we get to pick only one of three possibilities listed above?
Empty or the entire set. So, yes.

3. Originally Posted by novice
Let $R$ be an equivalence relation on $P(X)$ by $ARB$ if $A\cup Y=B \cup Y$, where $X=\{1,2,3,4\}$, and $Y=\{3,4\}$.
What is the equivalence class of $\{1,2\}$?

I know there is only one element of the power set $X$ for the answer such that

$A \cup \{3,4\} = \{1,2\} \cup \{3,4\}=\{1,2,3,4\}$

For $A$, the possibilities are

$\{1,2\}$
$\{1,2,3\}$
$\{1,2,3,4\}$

Question: Since the intersection of any two equivalence classes is an empty set, does it mean that we get to pick only one of three possibilities listed above?
No.
The equivalence class in question is: {{1,2}, {1,2,3}, {1,2,4}, {1,2,3,4}}.
And this set is a member of the partition of P(X) induced by R.
I think it's important to understand that it's the partition that's a collection of pairwise disjoint sets.

It's a good exercise; stretch it a bit. What's the rank of R?

4. Originally Posted by Drexel28
Empty or the entire set. So, yes.
By definition, every element of X belongs to exactly one equivalence class, so it must not be empty.

5. Originally Posted by PiperAlpha167
No.
The equivalence class in question is: {{1,2}, {1,2,3}, {1,2,4}, {1,2,3,4}}.
And this set is a member of the partition of P(X) induced by R.<--Can't find this in my book.
I think it's important to understand that it's the partition that's a collection of pairwise disjoint sets.

It's a good exercise; stretch it a bit. What's the rank of R?
The set of the distinct equivalence classes is a partition of the set $X=\{1,2,3,4\}$.

Since a set of equivalence classes is a collection of pairwise disjoint, nonempty subset of $X$ whose union is $X$. None of $\{1,2\}, \{1,2,3\},\{1,2,4\}$, $\{1,2,3,4\}$ are pairwise disjoint subsets of $X$.

I have a hunch that we are to pick only one of those while these are the possible partitions of $X$:

$\{\{1,2\},\{3,4\}\}$
$\{\{1,2,3\},\{4\}\}$<--doubtful
$\{\{1,2,4\},\{3\}\}$<--doubtful
or
$
\{\{1,2,3,4\}\}
$
<--this too is doubtful

I have strong doubt of for those three since they are not pairwise disjoint with Y. There is also another reason for doubt, since Y={3,4} is a fixed subset of X.

I wonder if there is a book that gives clear illustration in this regard.

6. Originally Posted by novice
The set of the distinct equivalence classes is a partition of the set $X=\{1,2,3,4\}$.

Since a set of equivalence classes is a collection of pairwise disjoint, nonempty subset of $X$ whose union is $X$. None of $\{1,2\}, \{1,2,3\},\{1,2,4\}$, $\{1,2,3,4\}$ are pairwise disjoint subsets of $X$.

I have a hunch that we are to pick only one of those while these are the possible partitions of $X$:

$\{\{1,2\},\{3,4\}\}$
$\{\{1,2,3\},\{4\}\}$<--doubtful
$\{\{1,2,4\},\{3\}\}$<--doubtful
or
$
\{\{1,2,3,4\}\}
$
<--this too is doubtful

I have strong doubt of for those three since they are not pairwise disjoint with Y. There is also another reason for doubt, since Y={3,4} is a fixed subset of X.

I wonder if there is a book that gives clear illustration in this regard.
R is an equivalence relation on P(X), not on X.

The set of equivalence classes partition P(X), not X.

Each equivalence class is a set of sets of numbers from X, not a set of numbers from X.

The disjointness has to do with these sets of sets.

You are focusing your attention in the wrong place. You are looking at detail inside of X.
This exercise is a good test case for demonstrating an ability to abstract. Back out of X, and elevate yourself by looking into P(X) -- all 16 members.
I asked you about the rank of R. When you've gained a better understanding of this material, you'll see that the rank is four.

7. Originally Posted by PiperAlpha167
R is an equivalence relation on P(X), not on X.

The set of equivalence classes partition P(X), not X.

Each equivalence class is a set of sets of numbers from X, not a set of numbers from X.

The disjointness has to do with these sets of sets.

You are focusing your attention in the wrong place. You are looking at detail inside of X.
This exercise is a good test case for demonstrating an ability to abstract. Back out of X, and elevate yourself by looking into P(X) -- all 16 members.
I asked you about the rank of R. When you've gained a better understanding of this material, you'll see that the rank is four.
You are absolutely right. I got distracted by what I read in the book.
Now as to the rank of R, the material in my book made no mention of the subject of rank. For this, I need more reading beyond what my book can offer. It's a good thing for me to look into, for the answer to question.

Thanks for you time and a very valuable lesson. I'll catch up you later with the rank of R if I could get a hold of the definition. I got 4 equivalence classes. Is that what you mean by rank?

8. Originally Posted by novice
You are absolutely right. I got distracted by what I read in the book.
Now as to the rank of R, the material in my book made no mention of the subject of rank. For this, I need more reading beyond what my book can offer. It's a good thing for me to look into, for the answer to question.

Thanks for you time and a very valuable lesson. I'll catch up you later with the rank of R if I could get a hold of the definition. I got 4 equivalence classes. Is that what you mean by rank?
Yes.
The rank (or index) of an equivalence relation R on a set S is the number of equivalence classes that make up the partition
of A induced by R (essentially, the cardinality of the partition). If the partition is infinite, the rank is said to be infinite.
But, apparently these terms have become archaic. Not too much on the Internet -- "amazingly", not even at Wiki.
(I see I've unwittingly dated myself.)
Anyway, I suggest disregarding them.