Relations on Sets: Need help understanding a mistake

[WWCY]

Homework Statement

Suppose $R$ and $S$ are relations on a set $A$.

If $R$ and $S$ are transitive, is $R \cup S$ transitive? Why?

Homework Equations

The Attempt at a Solution

Suppose that $a$ is an arbitrarily but particularly picked element of $R \cup S$, then
$$a \in R \ \text{or} \ a \in S$$
By division into cases, suppose $a \in R$,
Suppose $b,c$ are elements of $R$, such that $(a, b) \in R$ and $(b,c) \in R$
By definition of $R$, $(a,c) \in R$

I then carried out the exact same steps for the case $a \in S$ and concluded that $R \cup S$ is transitive.

However, I also managed to find a counter example by defining $A = \{ a,b,c \}$, $R = \{ (a,b) \}$, $S = \{ (b,c) \}$, $R \cup S = \{ (a,b) , (b,c) \}$. The fact that $b \in R \cap S$ suggests to me that my division into cases wasn't done right, but aren't my steps analogous to the division into cases seen in problems with statement variables?

Thanks in advance for any assistance!

 

[PeroK]

Suppose that $a$ is an arbitrarily but particularly picked element of $R \cup S$, then
$$a \in R \ \text{or} \ a \in S$$
By division into cases, suppose $a \in R$,
Suppose $b,c$ are elements of $R$, such that $(a, b) \in R$ and $(b,c) \in R$
By definition of $R$, $(a,c) \in R$

Just because $a \in R$ doesn't allow you to restrict yourself to $R$ for any remaining arbitrary elements of $R \cup S$.

I'm not familiar enough with the specific material to help further, but aren't relations a subset of the Cartesian product of a set with itself?

 

[PeroK]

PS Your counterexample is valid. Your proof is invalid because a) it treats relations as subset of $A$; and b) the mistake that without loss of generality you can take all elements to be in $R$.

 

[fresh_42]

If $R$ and $S$ are transitive, is $R \cup S$ transitive? Why?

If $(x,y)\in R\cup S$ then it is in either of them, say in $R$. Then for any $a\in A$ we find a path $(x,y) \sim \ldots \sim (z,a)$ in $R \subseteq R\cup S$.