Proving that a set relation is an equivalence relation

Question

I'm solving an exercise about equivalence relations. I have to show that this relation is an equivalent relation:

We define in $\mathbb{R}^{2}$ the relation given by $(x,y)R(a,b)\iff > \exists n\in \mathbb{Z}(n-1<y\leq n \wedge n-1<b\leq n)$

To show that it is an equivalent relation I have to prove that it is Reflexive, Symmetric and Transitive. I don't know how to prove that it is transitive.

For reflection: $\exists n\in \mathbb{Z}(n-1<y\leq n \wedge n-1<y\leq n)$ is always true.

For symmetry: $\exists n\in \mathbb{Z}(n-1<y\leq n \wedge n-1<b\leq n) \iff \exists n\in \mathbb{Z}(n-1<b\leq n \wedge n-1<y\leq n)$ is always true because the and operator is conmutative.

Thanks in advance!

Answer 1

Hint: The relation is of the form $\alpha R\beta\iff f(\alpha)=f(\beta)$. Here $f\colon \Bbb R^2\to\Bbb Z$, $(x,y)\mapsto \lceil y\rceil$

Answer 2

Suppose you have $(x,y)R(a,b)$ and $(a,b)R(e,f)$. Then you have

$\exists n\in \mathbb{Z}(n-1<y\leq n \wedge n-1<b\leq n)$

and

$\exists m\in \mathbb{Z}(m-1<b\leq m \wedge m-1<f\leq m)$

In particular you have $∃n∈Z(n−1<b≤n)$ and $∃m∈Z(m−1<b≤m)$. Try to show that you must have $m$ = $n$.

Answer 3

Hint:

If $f:X\to Y$ is a function and $a\sim b\iff f(a)=f(b)$ then $\sim$ is an equivalence relation on $X$.

Now see the hint of Hagen.

If it comes to e.g. symmetry then is simply enough to observe that:$$f(a)=f(b)\implies f(b)=f(a)$$

Proving that the relation is reflexive and transitive is of the same level.