Don't get why this is a transitive relation

Question

A = {(1,1), (2,2), (3,3)}.

So I know that this is a reflexive, antisymmetric and symmetric. But apparently it is also transitive, but I don't know how. Can anyone explain why it is transitive?

Answer 1

Transitivity means that if $aRb$ and $bRc$, then $aRc$. The implication $P\Rightarrow Q$ is true, if $P,Q$ are true or $P$ is false. Only in the case $P$ true, $Q$ false it is false.

Here you just can write $1R1$ and $1R1$ implies $1R1$.

In the example $\{(1,2),(2,3)\}$, the relation is not transitive, since $1R2$ and $2R3$, but not $1R3$.

Answer 2

For a transitive relation if we have $(x,y)$ and $(y,z)$ then we can conclude $(x,z)$ is also in the relation. In this case the only possible choice for $(x,y)$ and $(y,z)$ are of the form $(x,x)$ and sure enough $(x,x)$ is in the relation so it is transitive.

Answer 3

A relation $R$ fails to be transitive only if you can find $a,b,c$ such that $(a,b)\in R$ and $(b,c)\in R$ are true but $(a,c)\in R$ is not true.

Your relation is transitive since you can not find such $a,b,c$ to contradict transitivity.

Answer 4

This is just the equality relation on the set $S:=\{1,2,3\}$, so that $(a,b)\in A$ iff $a=b$, for any elements $a, b$ of the base set.
Of course, equality is transitive: $$(a=b) \land (b=c) \implies (a=c) $$

Answer 5

A relation $A$ over set $B$ is transitive if $\forall (x,y,z){\in}B^3~.((x,y){\in}A\wedge (y,z){\in}A\to (x,z){\in}A)$.

The negation of this is $\exists (x,y,z){\in}B^3~.((x,y){\in}A\wedge (y,z){\in}A\wedge (x,z){\notin}A))$.   Thus a relation is non-transitive only when it satisfies that existential statement.   The assignment of values of $x,y,z$ that witness this existence is called a counterexample.

So since there are not three such three pairs in the given $A$ that would satisfy that existential, then we cannot find a counterexample, and this relation is indeed transitive.