Let $A = \left \{ 1,2,3,4 \right \}$. Give an example of $R$ over $A$ so that it is symmetric and transitive but not reflexive.
My answer: $R = \begin{Bmatrix} (2,1)(1,2)(2,3)(1,3) \end{Bmatrix}$
Correct answer: $R = \begin{Bmatrix} (1,1)(2,2)(3,3) \end{Bmatrix}$
Question: Is my answer right? How is the ”correct” answer both transitive and symmetric?
On
The relation $R=\{(1,1)\}$ is symmetric, transitive but not reflexive. The same holds for any relation $R$ which is a proper subset of the diagonal $\{(1,1),(2,2),(3,3),(4,4)\}$.
On
Your answer is not transitive. You have $(2,1)$ and $(1,2)$ as elements in $S$, but not $(2,2)$ .
The correct answer is clearly symmetric because for every $(x,y)\in\{(1,1),(2,2),(3,3)\}$ then $(y,x)$ is too. It is transitive because there is no triple of $x,y,z$ where $(x,y)\in S,(y,z)\in S,(x,z)\notin S$.
It is not Reflexive because $(4,4)\notin S$. As @Wuestenfux pointed out while I was typing this, any proper subset of $\{(1,1),(2,2),(3,3),(4,4)\}$ will suffice, as will other sets, such as $\{(1,1),(1,2),(2,1),(2,2)\}$.
On
Your answer is not symmetric because $(2,3) \in S$ but $(3,2) \notin S$.
A relation $S$ is symmetric if $(a,b) \in S$ if and only if $(b,a) \in S$. Now, check the answer again and you can see that this symmetry condition is satisfied $(1,1) \in S$ and $(1,1) \in S$, etc.
A relation $S$ is transitive if $(a,b),(b,c) \in S$ then $(a,c) \in S$. In correct answer, you can see that there are also no two elements that breaks this condition (Note that $(1,1),(2,2) \in S$ does not even fit to the definition so it does not break the condition).
Your answer is not correct because it is not symmetric: $(1,3)$ is in $S$ but $(3,1)$ isn’t. The correct answer is symmetric $(a,b)\in S$ means $(b,a)\in S$ is trivially true, and transitive by the same triviality. You’re correct in saying that this question has many, many different solutions, however.