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Is the relation $R = \{(a,b),(a,c),(b,a),(b,c),(c,a),(c,b),(d,d)\}$ symmetric?

My professor claims that if $(d,d)$ was not included, it would be symmetric, but the inclusion of $(d,d)$ ruins it because $d$ has to connect to another element of the relation.

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  • $\begingroup$ Two of the defining properties of equivalence relations are that they are both reflexive and symmetric, which would be useless if they were mutually exclusive :-P $\endgroup$ Sep 24 '21 at 4:12
  • $\begingroup$ Hope your professor is willing to get corrected in this case! $\endgroup$ Sep 24 '21 at 8:39
  • $\begingroup$ You could tell your professor that $(d,d)\in R$ "connects" to itself, ie, for the element $(\color{red}d,\color{green}d)$ in $R$, we have the element $(\color{green}d,\color{red}d)$ in $R$. There is no restriction that it must "connect" to a distinct element in $R$. As an example, the relation $R=\{(a,a)\}$ on the singleton set $\{a\}$ is an equivalence relation: reflexive, symmetric and transitive. $\endgroup$ Sep 24 '21 at 19:16
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A very simple way to see if a relation $R$ is symmetric is to check its inverse relation $R^{-1}$, where $(x,y)\in R\Leftrightarrow (y,x)\in R^{-1}$

Since for the given relation, $R=R^{-1}$, it is symmetric.

$\color{green}{\text{YOU ARE CORRECT.}}$

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Yes, that relation is symmetric. The definition of symmetry does not require each element to be connected to some other element; $R$ is symmetric iff for every $x,y$ such that $(x,y)\in R$ it is also the case that $(y,x)\in R$.

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One way to see if a relation is symmetric is to draw the table what is in relation to what:

$$\begin{array}{r|c|c|c|c}R&a&b&c&d\\\hline a&F&T&T&F\\b&T&F&T&F\\c&T&T&F&F\\d&F&F&F&T\end{array}$$

and just see if the table is symmetrical across the main diagonal.

And - in this case it is. Thus, the relation is symmetric. Having $(d,d)$ in it has nothing to do with it, it would be symmetric either way (i.e. whether the entry for $(d,d)$ is "true" or "false").

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