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On the topic relations, is there any set in a relation R on $$ A = \{0,1,2,\{3\},4\} $$ which is reflexive, not symmetrical, not antisymmetrical and transitive? If a relation is transitive, shouldn't it also be at least antisymmetrical? For instance if the set is $$R = \{(0,1),(1,2),(0,2)\}$$ R should be transitive because $$(0,1) \land(1,2) \implies (0,2)$$ If R is transitive it should also be at least antisymmetrical, because $$(0,1)\land(1,0)\implies 1=0$$ The implication should be true because $$(1,0)\not\in R$$ $$ 1 \land0\implies0$$ $$ (0\implies 0) \Longleftrightarrow 1$$ is a true implication.

Is something wrong in my logical understanding on this topic? Am I missing something, I am not quite sure, if I really understood it correctly.

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    $\begingroup$ Are you sure about your set $A$? A set with elements $1, 2, 4 \in \mathbb N$, and the element, a set, $\{3\}$? $\endgroup$ – Namaste Oct 27 '18 at 19:58
  • $\begingroup$ To be reflexive, you'd need $R$ to include $(1, 1), (2, 2), (\{3\}, \{3\}),$ and $(4, 4)$. $\endgroup$ – Namaste Oct 27 '18 at 20:00
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    $\begingroup$ The four properties are independent of each other. Antisymmetrical allows only one-way paths between ANY different nodes. Symmetrical insists on "no path between a,b or two-way path". Again, this holds for ANY a,b. Mohammad has given the smallest convenient relation (unless you want to have $\{3\}$ and not $3$ :) $\endgroup$ – user376343 Oct 27 '18 at 20:00
  • $\begingroup$ yes, I am 100% sure. It is part of a set of exercises on the topic $\endgroup$ – Thomas Christopher Davies Oct 27 '18 at 20:01
  • $\begingroup$ Great. Just wanted to confirm, because the given answer is then incorrect. $\endgroup$ – Namaste Oct 27 '18 at 20:01
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Try this $$R = \{(0,0),(1,1),(2,2), (\{3\},\{3\}), (4,4),(1,2),(1,\{3\}),(\{3\},1)\}$$

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  • $\begingroup$ The set is not on which $R$ is a relation appears not to be $\{1, 2, 3, 4, \}$, which is the set on which your relation is built. The asker's set is $\{1, 2, \{3\}, 4\}$, unless the asker made a mistake in giving the set, which may be the case, $3 \notin A$. $\endgroup$ – Namaste Oct 27 '18 at 19:54
  • $\begingroup$ The set is as given with the $$\{3\}\in A$$ $\endgroup$ – Thomas Christopher Davies Oct 27 '18 at 19:57
  • $\begingroup$ can you give a little more detail on your answer? $\endgroup$ – Thomas Christopher Davies Oct 27 '18 at 19:58
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    $\begingroup$ @amWhy but it doesn't matter. Any set with 4 elements will do. $\endgroup$ – Jakobian Oct 27 '18 at 20:07
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    $\begingroup$ @ThomasChristopherDavies Thanks for the edit. $\endgroup$ – Mohammad Riazi-Kermani Oct 27 '18 at 20:56

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