TITLE: Is family of sets bounded from above an equivalence class of certain equivalence relation?
QUESTION [0 upvotes]: The equivalence relation $R$ on $\mathcal{P(\mathbb{R})}$ is given by 
$$ ARB \iff (\exists a \in \mathbb{R})A \cap [a, +\infty) = B \cap [a, +\infty)$$
and by family of sets bounded from above I mean 
$$\mathcal{A} = \{ C \subseteq \mathbb{R}: (\exists M \in R)(\forall x \in C) x \leq M \}$$
To me it seems that indeed $\mathcal{A}$ is an equivalence class of $R$, because I have not found a set that is equivalent to set bounded from above in this relation that is not bounded from above. But I have issues with justifying that statement. How can it be showed?

REPLY [1 votes]: As Don Thousand already points out in the comments, all elements of $\mathcal{A}$ are equivalent (under $R$). Just repeating the argument here: given any two sets $A$ and $B$ bounded above, there is an upper bound $M$ for both of them. Then $A \cap [M+1, \infty) = B \cap [M+1, \infty) = \emptyset$.
To see that $\mathcal{A}$ is indeed an equivalence class, we have to show that there is no set that is not in $\mathcal{A}$, which is related to something in $\mathcal{A}$. Let $B \subseteq \mathbb{R}$, $B \not \in \mathcal{A}$. For any $A \in \mathcal{A}$ we have some upper bound $M$, so $A \cap [M+1, \infty) = \emptyset$. Since $B \not \in \mathcal{A}$, we must have that $B$ is unbounded and thus $B \cap [M+1, \infty) \neq \emptyset$. So we see that $B$ is not related to anything in $\mathcal{A}$, and so $\mathcal{A}$ is indeed an $R$-equivalence class.
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