Is the relation $(a,A)R(b,B)\iff A\cup[0;b]=B\cup[0;a]$ transitive?

Question

Let $R$ be a relation on $\mathbb{N}\times\text{power set}(\mathbb{N})$, defined as $$(a,A)R(b,B)\iff A\cup[0;b]=B\cup[0;a]$$ Is $R$ transitive?

As the notation $[0,x]$ caused confusion in one of the answers, let me clarify that this means all real numbers between $0$ and $x$, inclusive, so all $y:0\le y\le x.$

Let $(a,A)R(b,B)R(c,C)$, so $A\cup [0;b]=B\cup [0;a]$ and $B\cup[0;c]=C\cup[0;b]$.

The question is if $$A\cup[0;c]=C\cup[0;a]?$$

I really don't know how to say if this equality holds, or not, so I decided to try to construct a counterexample, but I didn't succeed. I thought something like this $$(5, \mathbb{N}_{>5})R(5,\mathbb{N}_{>1})R(0,\emptyset),$$ where $\mathbb{N}_{>t}$ denotes the integer numbers greater that $t$. Well, the issue is I can't find such three tuples that the relationship holds between consecutive ones, but it doesn't between the first and third.

I literally cannot come up with a counterexample... Two ordered pairs being in the relation, according to me, means that their first components must definitely be equal. $[0, x]$ is a set that contains all real numbers between $0$ and $x$, inclusive. There is no way to compensate the real numbers from this set with any subset of $\mathbb{N}$.

May someone provide me with a motivation for finding a counterexample or proving that the relation is transitive? Thanks!

Answer 1

We seek to show that $(A, a)R(B, b) \wedge (B, b)R(C, c) \implies (A, a)R(C, c)$. you should easily see that $(A, a)R(B, b) \wedge (B, b)R(C, c) \implies a = b = c$ (otherwise you would be missing real numbers and one of the unions would be incorrecect). The answer is now trivially true, if you look at the tuples $(A, a), (B, a), (C, a)$.

You should note that it is not necessary that $A = B = C$ as you write, rather if $a = [0, a]$ then $ A - a = B - a = C - a$. This fact however is unnecessary for the proof just a minor correction to aid in your understanding :)

Answer 2

First note that if $(a,A)R(b,B)$ then the set of limit points of $A\cup[0,b]$, i.e. the set $(0,b)$, is equal to $(0,a)$, hence $b=a$.

Secondly, $A\cup[0,a]=B\cup[0,a]$ iff $A\cap[a+1,\infty)=B\cap[a+1,\infty).$

Using these two remarks, it is now clear that $R$ is an equivalence relation.

Answer 3

Let's suppose we have $(a,A)R(b,B)$ so $A\cup [0,b]=B\cup [0,a]$. If we assume $a< b$ then $(a,b)\subset B$. But as $(a,b)$ contains irrational numbers while $B$ contains only natural numbers this is impossible. [and the same holds if $a> b$]

So $(a,A)R(b,B) \implies a=b$.

So $(a,A)R(b,B)\iff a=b$ and $(a,A)R(a,B)\iff A\cup[0,a] =B\cup [0, a)\iff A\setminus [0,a]=B\setminus [0,a]$.

That's easy to show is equivalence relationship.

If $(a,A)R(b,B)$ and $(b,B)R(c,C)$ then $a=b$ and $b = c$ and $A\setminus [0,b] = B\setminus [0,a]= B\setminus [0,b]$ and $B\setminus[0,b]=B\setminus [0,c]=C\setminus[0,b]$ .... NOW I can go back to my claim the equality is transitive.