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Given the Binary Relation $R$

$R = \left \{ \left ( x,y \right ) | x = y+1\ and\ x,y \in \left \{ 0,1,2,3... \right \} \right \} $

The symmetric closure of $R$ is ?

I have the idea of symmetric relation, but what is the symmetric closure of $R$ here ?

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The symmetric closure of a relation $R$ is the intersection of all symmetric relations containing $R$. It is the "most parsimonious way to extend $R$ so it becomes symmetric". It also admits an easy expression: it is $R \cup \hat{R}$, where $\hat{R}$ is the relation which contains $(x,y)$ iff $(y,x)$ is in $R$.

For example, your $R$ has $0R1$, so its symmetric closure $\bar{R}$ must also have $1 \bar{R} 0$. However, it is not the case that $0 R 2$ or $2 R 0$, so it's not the case that $0 \bar{R} 2$.

For $R$ specifically, we have $$\hat{R} = \{(x,y) : (y,x) \in R \} = \{(x,y): y = x+1 \} = \{(x,x+1) : x \in \mathbb{N} \}$$ We also have $$R = \{(y+1,y): y \in \mathbb{N} \}$$ so the symmetric closure is $$R \cup \hat{R} = \{(x, x\pm1): x \in \mathbb{N}^{\geq 1} \} \cup \{(0,1)\}$$

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    Then, What is the representation in set form ? – 2017-01-27
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    You want $R$'s symmetric relation explicitly? – 2017-01-27
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    Yes, in set form ? – 2017-01-27
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    My professor says that $R = \left \{ \left ( x,y \right ) | y = x+/-1\ and\ x,y \in \left \{ 0,1,2,3... \right \} \right \} $ . – 2017-01-27
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    Your professor is correct. The LaTeX symbol \pm produces $\pm$. – 2017-01-27
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    Okk ! Thanks :-) – 2017-01-27