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$A = \{1,2,\dots,n\}$, and $R$ be a relation on $A$ (so that $R \subset A \times A$).

i) Is it possible to have an equivalence relation with exactly $n+1$ elements on $A$? If yes, give an example, if no, explain why?

For this I considered what adding an extra element from the possible relations would do. For example from $R = \{(1,1),(2,2),(3,3)\}$ adding an extra element $(1,2)$ would stop the relation from being symmetric, meaning adding in just one element would prevent this from being an equivalence relation. So you cannot have an equivalence relation with exactly $n+1$ elements in.

ii) Is it possible to have an equivalence relation with exactly $n+2$ elements on $A$? If yes, give an example, if no, explain why?

My reasoning for this question follows a similar patter to I, two elements could be added that allow it maintain being an equivalence relation. Adding in $(1,2)$ and $(2,1)$ would ensure it was still symmetric, and so still an equivalence relation.

iii) When $n = 3$, give all possible equivalence relations on $A$.

For this question I considered what adding in specific extra elements would require to be included in the final set.

  1. $(1,1), (2,2), (3,3)$
  2. $(1,1), (1,2), (2,1), (2,2), (3,3)$
  3. $(1,1), (1,3), (2,2), (3,1), (3,3)$
  4. $(1,1), (2,2), (2,3), (3,2), (3,3)$
  5. $(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)$

Are my answers and reasoning correct?

Also, should $\emptyset$ be included as an equivalency relation for III?

pc799
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    Well, to generalize your approach, since we must have the $n$ elements $(k,k)$ for each $k$, a new element would have to be of the form $(i,j)$ for $i\neq j$. But then we need $(j,i)$ as well, hence at least $n+2$ elements. – lulu Jan 18 '22 at 17:09
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    Your reasoning on the first two parts look good to me. When trying to list all the possible equivalence relationships on (say) ${1,2,3}$, it might be helpful to see that an equivalence relation amounts to a partition of the set. – hardmath Jan 18 '22 at 17:10
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    Looks fine although I would simply write 5. as $A^2$ rather than listing all of the elements. – CyclotomicField Jan 18 '22 at 17:11
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    i) Your reasoning on the first part is fine but you need to make it more general. What if $n > 3$ and what if the element you are adding isn't $(1,2)$. ii) ditto but how do you know it is transitive. – fleablood Jan 18 '22 at 17:16
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    Looks correct, all three. – Wuestenfux Jan 18 '22 at 17:18
  • @fleablood - I thought for Equivalency relations you would assume something is true during proof? There aren't any relations that I could add in that would be symmetric and allow me to disprove transitivity. I.e., I couldn't add in $(1,3),(2,1)$ since this would prevent it from being symmetric although this would allow me to disprove transitivity. Since the aim of the question is to prove it is possible, it would be counterintuitive to include elements that prevent symmetry. – pc799 Jan 18 '22 at 17:24
  • The empty set is not a equivalence relation on ${1,2,3}$. By reflexivity, $(1,1)$, $(2,2)$ and $(3,3)$ are required to be in the equivalence relation. – jjagmath Jan 18 '22 at 18:05
  • I meant, how do you know adding $(1,2)$ and $(2,1)$ doesn't break transitivity... "There aren't any relations that I could add in that would be symmetric and allow me to disprove transitivity" How do you know that? (Actually that is true, but how do you know that?) – fleablood Jan 18 '22 at 18:05
  • @fleablood - I understand your question, but I'm not entirely sure how I would go about answering it properly. Would it follow the lines of showing there is $(1,2)$ and $(2,1)$ and $(2,2)$ and since $(1,2)$ again it is transitive? I know that a relation is not transitive if $(a,b)$ & $(b,c)$ and there is no $(a,c)$ then it would not be transitive. Could I prove that since there is no $(b,c)$ there would be no $(a,c)$. – pc799 Jan 18 '22 at 18:20
  • @jjagmath - Thank you, I was just wondering since $\emptyset$ would be a proper subset of the relation $R$ if it could be an equivalence relation on its own. Is there a proof I could see for why $\emptyset$ would not be an equivalence relation? I was just thinking that if $R$ had nothing in at all then it could meet all the requirements for an equivalence relation. – pc799 Jan 18 '22 at 18:25
  • The proof that $\emptyset$ is not a equivalent relation on ${1,2,3}$ is straightforward. The definition of equivalent relation is that it must be reflexive, symmetric and transitive. But $\emptyset$ is not reflexive. Reflexive property says: For all $x \in A$, $(x,x) \in R$. The $\emptyset$ IS an equivalent relation on the set $A=\emptyset$. – jjagmath Jan 18 '22 at 19:07
  • We could note that if $(a,b), (b,c)\in R$ and $b\ne1$ or $2$ we must, in order to have $(a,b)\in R$ have $a=b$ and to have $(b,c)\in R$ we must have $b=c$. So $(a,c)=(b,b)\in R$. And if $b = 1$ or $2$ then to have $(a,b)\in R$ we must have $a=1$ or $2$ and to have $(b,c)\in R$ we must have $c=1$ or $2$. So $(a,c)$ consists of only $1$ or $2$ and is therefore in $R$. – fleablood Jan 18 '22 at 19:45

1 Answers1

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Recall that $A = \{1,\ldots,n\}$ is by construction not empty.

You added:

Also, should ∅ be included as an equivalency relation for III?

No. While the empty set is a relation (a subset of $A\times A$), it is not an equivalence relation. It is not reflexive.

The observation in my earlier Comment, that an equivalence relation on $A$ corresponds to a partition of the set $A$, has been discussed here before. We can use this idea to give a more formal justification of your ideas for parts 1,2,3.

By definition each of the $n$ elements of $A$ belong to one and only one equivalence class that appears in the corresponding partition of $A$.

If we are interested in counting the number of ordered pairs in such equivalence relations, it can be determined from the corresponding sizes of the equivalence classes that partition $A$, namely the integer partitions of $n = |A|$.

The smallest case is $n = 1 + 1 + \ldots + 1$ corresponding to the equality relation on $A$, i.e. using singleton subsets to partition $A$. The size of the equivalence relation is the sum of squares of the integer parts:

$$ 1^2 + 1^2 + \ldots + 1^2 = n $$

Since any other equivalence relation will have more relations, we can solve part (1) by observing that putting any two elements of $A$ into the same equivalence class would create four ordered pairs in place of the two that merely relate those two elements to themselves:

$$ 2^2 + 1^2 + \ldots + 1^2 = 4 + (n-2) = n + 2 $$

This not only answers (1) being impossible, it shows (2) is always possible (as long as $A$ contains more than one element).

hardmath
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