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