Does anybody know about a reflexive, antisymmetric, but not transitive relation on $\mathbb Z$? I really cant figure any out and I am having doubts that something like that exists.
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Let $R \subseteq \mathbb Z^2$ denote the relation to find, if we want $R$ to be reflexive, we need $$ \Delta_{\mathbb Z} = \{(z,z)\mid z \in \mathbb Z\} \subseteq R $$ $R$ is allready antisymmetric, we just must pay attention not destroying this property. Now for the not-transtivity: Add two points, say $(0,1), (1,2) \in R$, but $(0,2) \not \in R$. As $R := \Delta_{\mathbb Z}\cup\{(0,1), (1,2)\}$ is antisymmetric, we are done.
Addendum. If you want to have a formula, this can be done, our $R$ is given by $$ x\mathbin R y \iff (x-y)^2\cdot\bigl(x^2+(y-1)^2\bigr)\cdot \bigl((x-1)^2+(y-2)^2\bigr) = 0. $$
martini
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Great, thanks it enlights the think much more for me. However, I was maybe a bit unspecific, i meant a relation as a prescription or a formula. – Oskar Löw Oct 16 '13 at 12:59
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@OskarLöw See above. – martini Oct 16 '13 at 13:13