I am unsure how to find the all the equivalence relations. I know that a relation is an equivalence relation if it is reflexive, symmetric, and transitive. However I'm still uncertain how to find them all.
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1If you want to list all the equivalence relations, then you should be writing a list of sets of pairs of elements of $A$. Right now, I only see one set of pairs of elements of $A$. – Apr 16 '20 at 23:47
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I thought I wrote out all the pairs. What pairs am I missing – wcr221 Apr 16 '20 at 23:49
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1"List of sets of pairs" – Apr 16 '20 at 23:50
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so would it be what I wrote up there plus {(1,1)} , {(2,2)}, {(3,3)} – wcr221 Apr 16 '20 at 23:57
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Those three additional relations you wrote are not equivalence relations on $A$. – Apr 17 '20 at 00:13
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1It may help to note that there is a bijection between the set of equivalence relations over a set and the set of partitions over a set. The trivial equivalence relation where everything is related to everything, that is, $1\simeq 2\simeq 3$, corresponds to the partition ${{1,2,3}}$. The other trivial equivalence relation ${(1,1),(2,2),(3,3)}$, that is elements are only related to themselves, corresponds to the partition ${{1},{2},{3}}$. There are still three more partitions... the one where $1=2\neq 3$, the one where $1=3\neq 2$, and the one where $1\neq 2=3$ – JMoravitz Apr 17 '20 at 00:19
2 Answers
Let us recall the definition of an equivalence relation $R$ on a set $S$. It means the following hold:
- Reflexivity: for all $s \in S$, $(s,s) \in R$ (emphasis on for all)
- Symmetry: Whenever $(a,b) \in R$, so is $(b,a)$
- Transitivity: Whenever $(a,b),(b,c) \in R$, so is $(a,c)$
Bear in mind that the first condition requires that $(a,a),(b,b),(c,c) \in R$ for all equivalence relations $R$, so the "minimal size" relation $R$ is
$$R = \{ (a,a),(b,b),(c,c) \}$$
unlike what you're suggesting in the comments elsewhere.
It might be easiest to visualize this exercise through a table:
$$\begin{array}{|c|c|c|c|} \hline & a & b & c \\ \hline a & & & \\ \hline b & & & \\ \hline c & & & \\ \hline \end{array}$$
Put a mark in a given square. Then it will mean that $(x,y)$ is in the relation $R$, where $x$ is the element from the left and $y$ the element from above.
To be an equivalence relation, we need to be reflexive, transitive, and symmetric. Reflexive would require a mark on all elements of the diagonal as so:
$$\begin{array}{|c|c|c|c|} \hline & a & b & c \\ \hline a & X & & \\ \hline b & & X & \\ \hline c & & & X \\ \hline \end{array}$$
What else? Well, this already gives one equivalence relation, trivially. Suppose we add more pairs to $R$ that are non-diagonal entries. Then we would have to add the pair that is flipped across the diagonal, sort of like taking the transpose of a matrix. That means if you mark $(a,b)$ you have to mark $(b,a)$ too, and vice versa.
Just be sure to mark off further pairs that transitivity also has to imply!
Transitivity also has a condition related to this table method, but we instead have to use a condition that is slightly more formal and perhaps harder to visualize. Relations can be envisioned as matrices, where the rows and columns represent the elements, $0$'s are in the unrelated entries, and $1$ in the related entries. That is, for example,
$$\begin{array}{|c|c|c|c|} \hline & a & b & c \\ \hline a & X & & \\ \hline b & & X & \\ \hline c & & & X \\ \hline \end{array} \implies R = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]$$
In this case, the conditions are
- Reflexivity: All entries in the diagonal of $R$ are nonzero.
- Symmetry: $R = R^T$.
- Transitivity: Find $R^2$. If the nonzero entries in $R^2$ and those in $R$ are in the same position, we have transitivity.
Finally, as pointed out, just be sure you enumerate all of the equivalence relations. You only found one, but as you have already seen above, there's at least one you forgot.
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This is correct, but I think not a good approach for beginners. Using the fact that equivalence relations are just partitions is better than using the square of the matrix to check transitivity. – Ethan Bolker Apr 17 '20 at 00:26
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I think this might be it now but i'm still not sure if this is all of them { (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3) } and {(1,1), (2, 2), (3, 3)} and { (1,1), (1,2), (2,1), (2,2), (2,3), (3,2), (3,3)} and { (1,1), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3) } and { (1,1), (1,2), (1,3), (2,1), (2,3), (3,1), (3,3) } – wcr221 Apr 17 '20 at 00:55
That is one of the equivalence relations on the set: the relation in which everything is equivalent to everything else. There's another in which the only pairs are $(x,x)$: each element is related only to itself.
To find all the others, think about or look up the connection between equivalence relations and partitions. Here's one reference:
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What you wrote plus ${(1,1),(2,2),(3,3)}$ plus some others. Read the reference I suggested. – Ethan Bolker Apr 17 '20 at 00:24