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For each of the eight subsets of $\{ \mathrm{reflexive, symmetric, transitive} \}$, find a relation on $\{1,2,3\}$ that has the properties in that subset, but not the properties that are not in the subset.

Here is what I believe works for each. Let us denote the $8$ subsets by $1 = \emptyset, 2= \{r\}, 3 = \{s\}, 4 = \{t\}, 5 = \{r,s\}, 6 = \{r,t\}, 7 = \{s,t\}, 8 = \{r,s,t\}$.

Let $\rho_i$ be a relation with the properties of set $i$.

Then $\rho_1 = \emptyset$, $\rho_2 = \{(1,1)\}$, $\rho_3 = \{(1,2),(2,1)\}$, $\rho_4 = \{(1,2),(2,3),(1,3)\}$, $\rho_5 = \{(1,1), (1,2), (2,1)\} = \rho_1 \cup \rho_2$, $\rho_6 = \{(1,1), (1,2), (2,3), (1,3)\} = \rho_1 \cup \rho_4$, $\rho_7 = \{(1,2),(2,1), (2,3),(1,3)\} = \rho_3 \cup \rho_4$, and $\rho_8 = \{(1,1), (1,2), (2,1), (2,3), (1,3), (3,2), (3,1)\}$, the last one, somewhat surprisingly to me, is not $\rho_2 \cup \rho_3 \cup \rho_4$. My question is, do these $\rho_i$ work? For example in the only reflexive set I do not have $(2,2)$, but I believe this is fine, is it?

Martin Sleziak
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1 Answers1

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  1. $\rho_1$ is actually symmetric and transitive. Both properties hold vacuously as no elements are related to each other.
  2. $\rho_2$ is not reflexive. You require $(2,2)$ and $(3,3)$ to be inside the relation as well. It is also symmetric and transitive.
  3. $\rho_3$ is good.
  4. $\rho_4$ is good.
  5. $\rho_5$, similar to $\rho_2$, requires $(2,2)$ and $(3,3)$ to be in the relation for it to be reflexive. It is also transitive.
  6. $\rho_6$ has similar issues on reflexivity.
  7. $\rho_7$ is not symmetric.
  8. $\rho_8$ has similar issues in reflexivity. For this, you can simply take the power set.

Can you work out the issues I've listed out?

Clement Yung
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  • I see, can you clarify one point for me then? Why is it not okay leave out $(2,2)$ in $\rho_2$ but is it okay to not include say $(2,3)$ and $(3,2)$ in $\rho_3$. I'm sure the answer is obvious, but I'm just not seeing it. –  May 09 '20 at 06:19
  • The definition of reflexive is the following: $$ \forall x \in X : (x,x) \in \rho $$ Here, $X = {1,2,3}$. Thus, to ensure that it's reflexive, we must make sure that for every $x \in X$, $(x,x) \in \rho$, so $(2,2)$ and $(3,3)$ must both be in $\rho$. – Clement Yung May 09 '20 at 06:42
  • On the other hand, the definition of symmetric is as follows: $$ \forall x,y \in X : (x,y) \in \rho \implies (y,x) \in \rho $$ This is not the same as the following, which appears to be what you're thinking of: $$ \forall x,y \in X : (x,y) \in \rho \wedge (y,x) \in \rho $$ Symmetric tells us that if $(x,y) \in \rho$, then $(y,x) \in \rho$ as well. However, it doesn't say that every $(x,y)$ and $(y,x)$ must both be in $\rho$, as that would be the second definition (which is not what symmetry means). – Clement Yung May 09 '20 at 06:42