Following the question Example of a minimal equivalence relation on $A$ containing $R$, I tried to solve the following problem.
Problem: Find the minimal equivalence relation containing each of the relations given below.
Let $S$ be the relation on $\mathbb{R}$ given by $xSy$ if and only if $x = 2y$, for all $x, y \in \mathbb{R}$.
Let $P$ be the set of all people, and let $R$ be the relation on $P$ given by $xRy$ if and only if $x$ likes $y$, for all $x, y \in \mathbb{R}$.
The relation $<$ on $\mathbb{R}$.
I’m not sure if my answer is right or not and I hope someone can check it and discuss it. (I’m gonna extend my reasoning for part $1.$ to see if it’s correct, for the other two parts, it will be something similar)
My solution:
It is straightforward to verify that $S$ is neither reflexive, symmetric nor transitive. So let’s add the needed “ingredients” to produce an equivalence relation. We want this new relation to be reflexive, so we will add the condition that $xSy$ if and only if $x=y$. We also want to have a relation that is symmetric. So we add the condition that $xSy$ if and only if $2x = y$. Although when we search for the transitive properties we find that $x$ and $y$ are related by $S$ if and only if the lines of the form $y = kx$ where $k \in \{\text{positive even number and their multiplicative inverse}\}$ (for the sake of simplicity, let’s denote this last set as $T$). So the minimal equivalence relation containing $S$ (denoted by $S_m$) would be defined as follows: $xS_{m}y$ if and only if $y = kx$ with $k \in T$, for all $x, y \in \mathbb{R}$.
We want an equivalence relation containing $R$. We need this new relation to be reflexive, that’s accomplished by taking all the people that likes themselves. We want this relation to be symmetric, so we want people that like each other. We also require this relation to be transitive, but in this case this last property follows right from the other two. So let $D \subseteq P$ denote the set of all people that likes themselves and each other. Then $R_{m} = D \times D$.
For this part, we know that $<$ is transitive. To be reflexive, we obtain $\leq$. To be symmetric, we obtain $\leq$ or $>$. So $x<_{m}y$ if and only if $x \leq y$ or $x > y$. By the trichotomy law, we deduce that $<_{m} = \mathbb{R} \times \mathbb{R}$.
I hope someone can check this and give me a feedback.
Thank you so much for your attention!
Your answer for $(3)$ is correct, but the other two need a bit of work.
In $(1)$ your set $T$ is too big: you need to replace your set $T$ with $\{2^n:n\in\Bbb Z\}$. (Nothing forces $\langle 1,6\rangle$ to be in $S_m$ for instance.) And it would be a good idea to go through the details of actually verifying that $\{\langle x,y\rangle:y=2^nx\text{ for some }n\in\Bbb Z\}$ really is an equivalence relation.
In $(2)$ you don’t get transitivity for free from reflexivity and symmetry: what if $x,y$, and $z$ are three distinct people, $x$ likes $y$, and $y$ likes $z$? Then $R_m$ has to contain $\langle x,z\rangle$, but it doesn’t get added when you add just enough to make the relation reflexive and symmetric. You need to take the transitive closure of what you have (and then verify that it’s still reflexive and symmetric). Or you could do it the other way round: take the transitive closure of $R$, then take the reflexive and symmetric closure of that and verify that the resulting relation is still transitive; the verification is a little easier with this approach.