Relations Symmetry and Transitivity

Question

Given the following Relations over the set $M := \{α, β, γ\}$

$R1 := \{(α, α), (α, β), (β, α), (β, β), (γ, γ)\}$

How is $R1$ transitive?
The condition for transitivity is
$(a,y)\in R1 \text{ and }(y,x) \in R1 \implies (a,x) \in R1$

And how is $R2$ not a partial order?
Partial order if thing is antisymmetric, reflexive and transitive.

Answer 1

$R1$ is transitive, because $\forall \alpha,\beta, \gamma \in M$ there is a transitive pairs following the definition of transitivity:

$(\alpha,\beta) \land (\beta,\alpha) \implies (\alpha,\alpha) \in R1 $

$(\alpha,\beta) \land (\beta,\beta) \implies (\alpha,\beta) \in R1 $

$(\beta,\alpha) \land (\alpha,\beta) \implies (\beta,\beta) \in R1 $

$(\beta,\alpha) \land (\alpha,\alpha) \implies (\beta,\alpha) \in R1$

$(\alpha,\alpha) \land (\alpha,\beta) \implies (\alpha,\beta) \in R1 $

$(\beta,\beta) \land (\beta,\alpha) \implies (\beta, \alpha) \in R1$

That proves that R1 is transitive.

R2 is not a partial order, because although it is reflexive it is both antisymmetric and symmetric. The requirements for partial order is that it should be antisymmetric, reflexive and transitive.

Antisymmetric because $(\alpha,\gamma) \in R2 \land (\gamma, \alpha) \notin R2$

Symmetric because $(\beta,\gamma) \in R2 \land (\gamma,\beta) \in R2$

Answer 2

Here is an illustration of $R_1$(with the usual conventions):

In this illustration, we can clearly see that $R_1$ is reflexive and symmetrical.

We have to check that $R_1$ is transitive:

$\forall (a,b,c)\in M^3, ((a,b)\in R_1\text{ and }(b,b)\in R_1)$ obviously implies that $(a,b)\in R_1$. Likewise for $((a,a)\in R_1\text{ and }(a,b)\in R_1)$

It therefore remains to be verified that $\color{red}(\color{blue}((\alpha, \beta) \in R_1 \land (\beta,\alpha)\in R_1\color{blue})$ and $\color{blue}((\alpha,\alpha)\in R_1 \land (\beta,\beta) \in R_1\color{blue})\color{red})$ is TRUE.

Probably the whole exercise was to check that $R_1$ is an equivalence relation on $M$.