# Thread: Equivalence Relation

1. ## Equivalence Relation

Give an example of an equivalence relation R on a set A = {1,2,3,4,5,6,7} with P the set of equivalence classes such that the following four properties are satisfied:
1. |P|=3
2. There exists no set S in P such that |S|=3
3. 3 R 4 but 3 R 5
4. there exists a set T in P such that 1, 7 in T

So far the only relations elements in the relation I can think of is:

R = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7)
(3,5), (5,3), (1,7), (7,1)}

I'm not sure how to get the rest... I seem to be missing a key part of how I am supposed to do this.

2. ## Re: Equivalence Relation

In general, you are looking for a partition of the elements of the original set into three sets, no set has exactly 3 elements, one sets contains both 3 and 5, one set (not necessarily different) contains 1,7. 4 is not in the set containing 3. Example:

$$\{1,2,3,5,7\}, \{4\}, \{6\}$$

This is a valid partition that satisfies all of the conditions. See if you can describe the relation that generates this partition.

3. ## Re: Equivalence Relation Originally Posted by MrJank Give an example of an equivalence relation R on a set A = {1,2,3,4,5,6,7} with P the set of equivalence classes such that the following four properties are satisfied:
1. |P|=3
2. There exists no set S in P such that |S|=3
3. 3 R 4 but 3 R 5
4. there exists a set T in P such that 1, 7 in T

So far the only relations elements in the relation I can think of is:

R = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7), (3,5), (5,3), (1,7), (7,1)}
I'm not sure how to get the rest... I seem to be missing a key part of how I am supposed to do this.
If we use $\mathcal{R}_x$ to stand the equivalence class determined by $x$ then that relation has five classes not three.
$\mathcal{R}_1=\{1,7\}$,$\mathcal{R}_2=\{2\}$,$\mathcal {R}_3=\{3,5\}$,$\mathcal{R}_4= \{4\}$,$\mathcal {R}_6= \{6\}$

4. ## Re: Equivalence Relation

R = {(1,1),(1,2),(1,3),(1,5),(1,7),(4,4),(6,6)}

Is that close?

5. ## Re: Equivalence Relation Originally Posted by MrJank R = {(1,1),(1,2),(1,3),(1,5),(1,7),(4,4),(6,6)}
Is that close?
That is not even an equivalence relation!
Say the classes are: $\mathcal{R}_1=\{1,7\},\mathcal{R}_2=\{2,3,5,6\}, \mathcal {R}_4=\{4\}$[/QUOTE]

Is that collection a partition of the set $\{1,2,3,4,5,6,7\}~?$
If it is what is the corresponding equivalence relation?