8.20\21 Relation Tutorial
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If \(\R\) is a relation on \(X\), the inverse of \(\R\) is the relation \(R^{-1}=\left\{(y,x):~_{x}R_y\right\}\)
Prove that \(\R\) is symmetric, if and only if, \(R=R^{-1}\)
Proof:
\(\Rightarrow\))Suppose that \(\R\) is symmetric, then \(_{x}R_{y}\Leftrightarrow_{y}R_{x}\Leftrightarrow~_{x}R^{-1}_{y}\)
Therefore \(R=R^{-1}\)
\(\Leftarrow\)) Suppose that \(R=R^{-1}\) then
\(_{x}R_{y}\Leftrightarrow_{x}R^{-1}_{y}\Leftrightarrow~_{y}R_{x}\). Therefore \(\R\) is symmetric.
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Let \(R\) be a relation on \(X\). The relation \(I_x=\{(x,x):~_xR_x\}\) is called the identity relation on \(X\).
Prove that \(R\) is antisymmetic, if and only if, \(R\cap R^{-1}\subseteq I_x\)
Proof
\(\Rightarrow\)) Suppose that \(R\) is antisymmetric then take element \((x,y)\in R\cap R^{-1}\Rightarrow(x,y)\in R~and~(x,y)\in R\Rightarrow ~_xR_y~and~_xR^{-1}y\Rightarrow ~_xR_y~and~_yRx\Rightarrow x=y\Rightarrow(x,y)=(x,x)\in I_x\)
Therefore \(R\cap R^{-1}\subseteq I_x\)
\(\Leftarrow\)) Suppose that \(R\cap R^{-1}\subseteq I_x\), that if \(~_xR_y~and~_yRx\Rightarrow ~_xRy~and~_xR^{-1}_y\Rightarrow (x,y)\in R~and ~(x,y)\in R^{-1}\Rightarrow(x,y)\in R\cap R^{-1}\subseteq I_x\Rightarrow x=y\)