Equivalence Relations: x~y if and only if {x,y} ⊂ X or {x,y} ⊂ (Y\X)

Question

Apologies if my title is incorrectly formatted. In my preview the MathJax commands don't seem to be working, so I've copied and pasted from other questions on SE.

The body of my problem:

We have two non-empty sets, X and Y, such as that X ⊊ Y. On X we define the equivalence relation xRy if and only if {x,y} ⊂ X or {x,y} ⊂ (Y\X). Prove that the relation is an equivalence relation and find all of its equivalence classes.

Now, I'm familiar with the definition of equivalence relation, however, I'm not sure how to test if the relation is symmetric or transitive in this example. I am also much less confident in my comprehension of equivalence classes, so if someone could explain how to find them in this example, I would be much appreciated.

Thank you SE.

Answer 1

Hint: To prove symmetry, use the fact that sets are unordered collections, such that $\{x,y\} = \{y,x\}$ (one way to justify this is with the Axiom of Extensionality). To prove transitivity, assume you have $x,y,z$ which satisfy $xRy$ and $yRz$. You wish to show that $xRz$, or in other words, that $x$ and $z$ either both belong to $X$ or both belong to $Y \setminus X$. To accomplish this, try showing that $x$, $y$, and $z$ all belong to $X$ or all belong to $Y \setminus X$.

Once you have proven that $R$ is an equivalence relation, recall that the equivalence class (with respect to equivalence relation $R$) of an element $x \in Y$ is $[x] = \{y \mid xRy\}$. Given $x \in Y$, consider two cases, one where $x \in X$ and another where $x \in Y \setminus X$, and try to find the equivalence class of $x$ in each case.

Answer 2

As you know, equivalence relations require three properties: $$\begin{align} &\mathrm{1. Reflexivity:}\quad x\sim x \\ &\mathrm{2. Symmetry:}\quad x\sim y \Leftrightarrow y\sim x \\ &\mathrm{3. Transativity:}\quad (x\sim y \wedge y\sim z)\Rightarrow x\sim z \end{align}$$

In your case:

1. Reflexivity holds because $\{x,x\}=\{x\}$

2. Symmetry holds because $\{x, y\}=\{y, x\}$, hence both $x\sim y$ and $y\sim x$.

3. Transativity holds because if $\{x, y\}\subset A \hspace{2px}\wedge \{y, z\}\subset A$ then $\{x, y, z\}\subset A \Rightarrow \{x, z\}\subset A$. $A$ can be either of $X$ or $(Y\setminus X)$.

The two equivalence classes are $X$ and $(Y\setminus X)$ because an element can either be in one or the other, and if two elements, $x$ and $y$, are in the same set, the set $\{x, y\}$ is a subset of the set, where $x$ and $y$ are, so they are equal.