Given a relation $R = \{(x,y)\mid y=x+1\}$ and I have to find the reflexive, transitive and symmetric closure.
For reflexive, I added $y=x$ with given condition so now the relation becomes
$R = \{(x, y)~| ~y=x+1 , \text{ or}\; y=x\}$
To make it symmetric, I added $y=x-1$ as one more condition but I am confused about transitive closure, what will be that?
Will the final result be a reflexive relation ?
Best Answer
So far, it seems you have $R=\{(x, y)\mid y=x+1, \text{ or}\; y=x \text{ or}\; y = x-1\}$.
For transitive closure, you need to have the relation satisfy, for all $x, y, z$, if $(x, y) \in R$ and $(y, z) \in R$, then your need to ensure $(x, z) \in R$.
Test your current conditions for the relation and check for transitivity.
For example, if $y = x + 1$ and $z = y + 1$, then $z = (x + 1) + 1 \implies z = x+2$. So currently, you don't have transitive closure. Both $(x, y), (y, z)\in R$, but $(x,z) \notin R$.
Note that $R = \{(x, y) \mid y = x\}$, by itself, is an equivalence relation (hence reflexive, symmetric, and transitive for all elements on which it is defined), as equality is perhaps the most fundamental of all equivalence relations.
Assuming we are talking about real numbers, we can get transitive closure (with reflexive closure), on the reals using the relation $R = \{(x, y)\mid y\leq x\}$. But this relation fails to by symmetric. "$\leq$" is a paradigmatic partial order relation on the set of reals: reflexive, antisymmetric, and, and transitive. Can you see why it is transitive, reflexive, but not symmetric?