Equivalence class $ a \mathcal{R}_\delta b \quad\text{if}\quad \frac{a+b}{ab} < \delta$

Question

For any fixed $\delta>0$, we define the relation $\mathcal{R}_\delta$ on the open real interval $(0,\infty)$ by $$ a \mathcal{R}_\delta b \quad\text{if}\quad \frac{a+b}{ab} < \delta. $$

Note that, if a relation is defined on a set $S$, then we can think about the corresponding relation on any subset $T$ of $S$, which agrees with the original relation everywhere on $T$. Although it is technically an abuse of notation, we use the same notation to denote both relations. We make it clear to which relation we refer by writing of the relation "on $S$" or the relation "on $T$".

For any $\delta>0$,

1) $\mathcal{R}_\delta$ is symmetric on $X$.

2) $\mathcal{R}_\delta$ is an equivalence relation on $Y$

3) $\mathcal{R}_\delta$ is transitive, but not an equivalence relation on $Z$.

For any fixed $\delta>0$, the intervals listed in proposition above are the largest real open intervals for which the proposition is true.

Answer 1

If I understand you right, you are looking for subsets $X$, $Y$ and $Z$ of $(0,+\infty)$ such that, for an arbitrary but fixed (before the choice of the sets) $\delta > 0$:

1. $\mathcal{R}_{\delta}$, restricted to $X$ is symmetric on $X$,
2. $\mathcal{R}_{\delta}$, restricted to $Y$ is an equivalence relation on $Y$,
3. $\mathcal{R}_{\delta}$, restricted to $Z$ is transitive on $Z$, but not an equivalence relation.

The observation that vrugtehagel made (that $\frac{a+b}{ab} = \frac{1}{a} + \frac{1}{b}$) is very useful.

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Fix $\delta > 0$.

First, notice that $$a \mathcal{R}_{\delta} b \Leftrightarrow \frac{1}{a} + \frac{1}{b} < \delta \Leftrightarrow \frac{1}{b} + \frac{1}{a} < \delta \Leftrightarrow b \mathcal{R}_{\delta} a.$$ So the relation is symmetric on any subset of $(0,+\infty)$ and $X$ may be any subset of the interval.

For the second problem, we need $\mathcal{R}_{\delta}$ to be reflexive too. Now $$a \mathcal{R}_{\delta} a \Leftrightarrow \frac{1}{a} + \frac{1}{a} < \delta \Leftrightarrow \frac{2}{a} < \delta \Leftrightarrow a > \frac{2}{\delta},$$ and so the relation is reflexive on any subset of $\left(\frac{2}{\delta},+\infty\right)$.
Likewise, you may easily see that for any $a, b \in \left(\frac{2}{\delta},+\infty\right)$, we have $a \mathcal{R}_{\delta} b$, and so the relation is certainly an equivalence on any subset of that interval (it relates every two elements of the set). Hence you may choose any $Y \subseteq \left(\frac{2}{\delta},+\infty\right)$.

Finally, you want a set $Z$ such that $\mathcal{R}_{\delta}$ is transitive on $Z$, but not an equivalence relation.
Let $Z = \{ \frac{2}{\delta} \}$.
By our reasoning above, $\mathcal{R}_{\delta}$ is not reflexive on $Z$ (in fact, it is empty), and so it is not an equivalence relation.
However, $\mathcal{R}_{\delta}$ is vacuously transitive on $Z$ (we cannot find $a,b,c \in Z$ such that $(a, b), (b,c) \in \mathcal{R}_{\delta}$, but $(a, c) \notin \mathcal{R}_{\delta}$).