Determining if a relation is reflexive, symmetric, or transitive

Question

Let $A = \{0,1,2,3\}$

Define a relation $T$ on $A$ as follows:

$T = \{(0,1),(2,3)\}$

Is $T$ reflexive? symmetric? transitive?

Answer 1

To be Reflexive you should have (0,0) (1,1) (2,2) (3,3)

To be symmetric if you have (0,1), then you should have (1,0)

To be transitive if you have (1,2) (2,4), then you should have (1,4)

To answer your question NO for Symmetric and Reflexive Yes for transitive because no counter example

Answer 2

A relation is reflexive $\iff$ $\forall_{\alpha}(\alpha \in A)$$(\langle\alpha,\alpha\rangle \in T)$. This is clearly not the case for T.

A relation is symmetric $\iff$ $\forall_{\alpha, \beta}(\alpha \in A)([\langle\alpha,\beta\rangle\in T] \rightarrow [\langle\beta,\alpha\rangle \in T])$. This isn't the case for T either.

A relation is transitive $\iff$ $\forall_{\alpha,\beta,\gamma}([\langle\alpha,\beta\rangle \in T \land \langle\beta,\gamma\rangle \in T] \rightarrow [\langle\alpha,\gamma\rangle \in T])$. Yes, this statement is vacuously true, I believe.

Answer 3

Are you familiar with the definitions of reflexive, symmetric and transitive relations?

A reflexive relation is a binary relation on a set for which every element is related to itself.

As you can clearly see $(0,0),(1,1)$ etc. are not contained in your relation, so it is not reflexive.

A relation is symmetric if $aRb \implies bRa$.

Once again, $(0,1)$ is there in your relation but $(1,0)$ isn't. So it is not symmetric.

Relation is transitive if $(aRb \wedge bRc) \implies aRc$.

There is no counter-example to this in your relation, so it is transitive.