- #1
evinda
Gold Member
MHB
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Hello! (Wave)
A relation [tex]R[/tex] on a set [tex]A[/tex] is called strict total order if it is total order and satisfies the trichotomous identity, i.e. if [tex]R[/tex] is:- antireflexive, so [tex](\forall x \in A) \lnot(xRx)[/tex] or [tex](\forall x \in A) <x,x> \notin R[/tex]
- antsymmetric, so [tex](\forall x \in A)(\forall y \in A) (xRy \rightarrow \lnot yRx)[/tex] or [tex](\forall x \in A) (\forall y \in A) (<x,y> \in R \rightarrow <y,x> \notin R)[/tex]
- transitive, so [tex](\forall x \in A) (\forall y \in A) (\forall z \in A)(x Ry \wedge yRz \rightarrow xRz)[/tex]
- [tex](\forall x \in A)(\forall y \in A) \rightarrow xRy \lor yRx \lor x=y[/tex]According to my notes, the relation of total order at the real numbers is a relation of strict total order.[tex]\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is smaller that y}\}[/tex]But, in this case it will never be [tex]x=y[/tex], right? (Thinking)
A relation [tex]R[/tex] on a set [tex]A[/tex] is called strict total order if it is total order and satisfies the trichotomous identity, i.e. if [tex]R[/tex] is:- antireflexive, so [tex](\forall x \in A) \lnot(xRx)[/tex] or [tex](\forall x \in A) <x,x> \notin R[/tex]
- antsymmetric, so [tex](\forall x \in A)(\forall y \in A) (xRy \rightarrow \lnot yRx)[/tex] or [tex](\forall x \in A) (\forall y \in A) (<x,y> \in R \rightarrow <y,x> \notin R)[/tex]
- transitive, so [tex](\forall x \in A) (\forall y \in A) (\forall z \in A)(x Ry \wedge yRz \rightarrow xRz)[/tex]
- [tex](\forall x \in A)(\forall y \in A) \rightarrow xRy \lor yRx \lor x=y[/tex]According to my notes, the relation of total order at the real numbers is a relation of strict total order.[tex]\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is smaller that y}\}[/tex]But, in this case it will never be [tex]x=y[/tex], right? (Thinking)