Empty closure of relation

Question

I have the following relation symbolized as $\sim$, defined on $\mathbb{Z}\times\mathbb{Z}$:

$$x\sim y \iff |x-y|=1$$

I would like to know how the transitive-reflexive closure looks like in this case, because this relation isn't transitive or reflexive, is it just an empty set?

In that case, it can't be an equivalence relation right?

Answer 1

Let's call $\sim_n$ the general relation $(x,y)\in\mathbb Z^2, x\sim_n y\iff |x-y|=n$.

Let's call $\thickapprox$ the reflexive-transitive closure of $\sim_1$.

To render $\sim_1$ reflexive, you are just interested by the domain of definition of this relation, this is $\mathbb Z$.

So we just say that $\forall x\in\mathbb Z, x\thickapprox x$.

There is no need for symetric closure, since $\sim_1$ is already symetric.

Nevertheless $\forall (x,y)\in\mathbb Z^2, x\thickapprox y\iff y\thickapprox x$

No, let's have a look at transitivity.

Let's call $E_n$ the subset of $\mathbb Z^2$ such that $(x,y)\in E_n\iff x\sim_n y$.

The first step of the transitive closure is to consider all elements of $U_0=E_1$ and make $\thickapprox$ transitive on this set.

So, if $(x,y),(y,z)\in (U_0)^2$ we want $(x\thickapprox y \text{ and } y\thickapprox z)\implies x\thickapprox z$.

But in $U_0=E_1,\quad x\thickapprox y$ is the same than $x\sim_1 y$ so

$|x-y|=1 \iff x=y+\epsilon_1$ where $\epsilon_1=\pm 1$

$|y-z|=1 \iff y=z+\epsilon_2$ where $\epsilon_2=\pm 1$

$x=z+\epsilon_1+\epsilon_2=z+\epsilon_3$ where $\epsilon_3\in\{-2,0,2\}$, thus $x\sim_0 z$ or $x\sim_2 z$.

Now we define $U_1=E_0\cup E_1\cup E_2$ and we have to make $\thickapprox$ transitive on this set.

I let you convince yourself that now $\epsilon_3=\epsilon_1+\epsilon_2$ can take the values $-4,-3,-2,-1,0,1,2,3,4$

The next set is $U_2=E_0\cup E_1\cup E_2\cup E_3\cup E_4$.

We have to continue the process until all possibilities are exhausted, and this lead to $U_\infty=\mathbb Z$.

Finally $\forall (x,y)\in\mathbb Z, x\thickapprox y\ $ and the closure of $\sim_1$ is not very interesting since there is only $1$ equivalence class.

Note 1: for $n\ge 2$ the closure of $\sim_n$ is a little more interesting, $x\thickapprox y$ if $x,y$ differ from a multiple of $n$, so there are $n$ equivalence classes.

Note 2: $\sim_0$ is already an equivalence relation, so it is its own closure, yet not very interesting either because $\mathbb Z/\sim_0=\mathbb Z$.

Note 3: I suppose you have now guessed that the closure of $\sim_n$ is the equality $\pmod{n}$.

Answer 2

The "transitive reflexive closure" $T$ of relation $R$ means one extends $R$ by including (for all $x,y,z$ which are related by $R$ at least once) all pairs $(x,x),$ and if $xRy$ include also $(y,x),$ and finally if both $xRy$ and $yRz$ include $(x,z)$ in $T.$

So in $|x-y|=1$ on $Z$ then $T$ would include all pairs $(x,y)$ where $x,y \in \mathbb{Z}.$