# Equivalence relation on set $\{0,1,2,3\}$

I'm given a a relation on the set above as $R = \{(0,0), (1,1), (2,2), (3,3)\}$. I can see how this is reflexive. Since if $a = 0,1,2, 3$ then $(a,a)\in R$. However, how is it symmetric and transitive? For it to be symmetric $(a,b)$ and $(b,a)$ have to be in $R$ and for transitive, for $(a,b), (b,c), (a,c)$ has to be in $R$. But I don't see for which values of $a,b,c$ the relational set is symmetric and transitive.

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Symmetric means if $(a,b) \in R$, then $(b,a) \in R$. In your example, all elements are of the form $(a,a)$ so it is true. Transitivity is simple, because in this case, if $(a,b),(b,c) \in R$, then you must have $a=b=c$. Hence it is trivially transitive. – copper.hat Nov 10 '12 at 16:54
You should make that an answer since it was a perfect response. – user1766888 Nov 10 '12 at 16:56
Took your suggestion, thanks! – copper.hat Nov 10 '12 at 16:57

Symmetric means if $(a,b) \in R$, then $(b,a) \in R$. In your example, all elements are of the form $(a,a)$ so it is true. Transitivity is simple, because in this case, if $(a,b),(b,c) \in R$, then you must have $a=b=c$. Hence it is trivially transitive.

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Hint

Use $(a,b)\in R \Leftrightarrow a = b$ .

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For it to be symmetric (a,b) and (b,a) have to be in the set and for transitive, for (a,b) and (b,c), (a,c) has to be in the relational set.

No, you have a serious misunderstanding here. Symmetry of $R$ just says that whenever an ordered pair belongs to $R$, its reversal also belongs to $R$. It says nothing about specific labels $a$ and $b$, and it does not require the components of the ordered pair to be distinct. If a relation $R$ is symmetric, and the pair $\langle\text{thing}_1,\text{thing}_2\rangle$ happens to belong to $R$, then you know that the pair $\langle\text{thing}_2,\text{thing}_1\rangle$ must also belong to $R$. The components $\text{thing}_1$ and $\text{thing}_2$ can be $a$ and $b$, $x$ and $y$, $b$ and $a$, $u$ and $u$, $\xi$ and $\eta$, $3$ and $5$, $17$ and $17$, or anything else, depending on the underlying set of the relation.

In your case every pair in the relation is its own reversal, so the relation is automatically symmetric: it’s automatically true that if a pair $\langle\text{thing}_1,\text{thing}_2\rangle$ is in $R$, the reversed pair $\langle\text{thing}_2,\text{thing}_1\rangle$ is also in $R$, because they’re the same pair.

To put it another way, the only way to demonstrate that a relation $R$ is not symmetric is to find objects $\text{thing}_1$ and $\text{thing}_2$ in its underlying set (which in your case is the set $\{0,1,2,3\}$) such that

$$\langle\text{thing}_1,\text{thing}_2\rangle\in R$$ but

$$\langle\text{thing}_2,\text{thing}_1\rangle\notin R\;.$$

And in your case that’s impossible.

You have a similar misunderstanding with respect to transitivity. Suppose that $R$ is a relation on a set $A$. Think of $A$ as a bunch of stepping-stones in a stream, and $R$ as a kind of list of the steps that are safe to take: if some pair $\langle x,y\rangle$ is in $R$, then you can step from $x$ to $y$. Transitivity of $R$ then says that if $x,y$, and $z$ are stepping-stones, not necessarily distinct, and if you can step from $x$ to $y$ and from $y$ to $z$, then you can step directly from $x$ to $z$. (No, it’s not realistic, but it may help explain the idea.)

In the case of your relation, if you can step from $x$ to $y$ and from $y$ to $z$, then $x=y=z$, because the only ‘steps’ that your $R$ allows are stepping in place, from $0$ to $0$, $1$ to $1$, $2$ to $2$, or $3$ to $3$. Thus, if you can step $x$ to $y$ and from $y$ to $z$, you can certainly step directly from $x$ to $z$: $x=y=z$ and the steps from $x$ to $y$, from $y$ to $z$, and from $x$ to $z$ are in fact all just the same step under different names.

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You have the set $X = \{0,1,2,3\}$. You define a relation by saying that for $x,y\in X$, $x\sim y$ if and only is $(x,y) \in R$. Now (as mentioned in another answer) this is the same as saying that $x \sim y$ if and only if $x = y$.

So it is symmetric because if $x\sim y$ then $x = y$ then $y =x$, and so $y\sim x$.

It is transitive because if $x\sim y$ and $y\sim z$ then $x = y$ and $y = z$ so $x = z$ and therefore $x \sim z$.

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