To demonstrate that $R_f$ is an equivalence relation and to describe the equivalence classes, we proceed with the following steps:

Part 1: Show that $R_f$ is an equivalence relation

An equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity. Let’s check these properties for $R_f$.

Reflexivity:

A relation $R$ on a set $X$ is reflexive if, for all $x \in X$, $(x, x) \in R$. We need to show that for every $x \in X$, $(x, x) \in R_f$.

By the definition of $R_f$, $(x, x) \in R_f \iff f(x) = f(x),$ which is true because a function is always equal to itself for the same input. Therefore, $R_f$ is reflexive.

Symmetry:

A relation $R$ on a set $X$ is symmetric if, for all $x, y \in X$, $(x, y) \in R$ implies $(y, x) \in R$. We need to show that if $(x, y) \in R_f$, then $(y, x) \in R_f$.

By the definition of $R_f$, $(x, y) \in R_f \iff f(x) = f(y).$ Since equality of functions is symmetric, $f(x) = f(y) \iff f(y) = f(x)$, we have that $(y, x) \in R_f$. Therefore, $R_f$ is symmetric.

Transitivity:

A relation $R$ on a set $X$ is transitive if, for all $x, y, z \in X$, $(x, y) \in R$ and $(y, z) \in R$ implies $(x, z) \in R$. We need to show that if $(x, y) \in R_f$ and $(y, z) \in R_f$, then $(x, z) \in R_f$.

By the definition of $R_f$, $(x, y) \in R_f \iff f(x) = f(y),$ and $(y, z) \in R_f \iff f(y) = f(z).$ If $f(x) = f(y)$ and $f(y) = f(z)$, then $f(x) = f(z)$, so $(x, z) \in R_f$. Therefore, $R_f$ is transitive.

Since $R_f$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Part 2: Define the equivalence classes

Given that $R_f$ is an equivalence relation, for each $x \in X$, the equivalence class of $x$, denoted by $[x]$, consists of all elements in $X$ that are related to $x$ under $R_f$. This means: $[x] = \{ y \in X \mid (x, y) \in R_f \}.$ By the definition of $R_f$, $(x, y) \in R_f \iff f(x) = f(y).$ Thus, the equivalence class of $x$ is: $[x] = \{ y \in X \mid f(x) = f(y) \}.$ In other words, the equivalence class of $x$ contains all elements of $X$ that are mapped to the same value as $x$ under the function $f$.

Alternatively, for all $x \in X$ and for every $y \in f(X)$ (where $f(X)$ is the image of $f$), the equivalence class of $x$ can be described as: $[x] = \{ x' \in X \mid f(x') = f(x) \}.$

Conclusion:

We have shown that $R_f$ is an equivalence relation because it satisfies reflexivity, symmetry, and transitivity. Moreover, the equivalence class of an element $x \in X$ under $R_f$ is the set of all elements in $X$ that are mapped to the same value as $x$ by the function $f$.

Do you need further clarification, or would you like more details on any of these points?

Here are five follow-up questions to deepen understanding:

1. How does the definition of an equivalence relation change if we modify the function $f$?
2. Can $R_f$ fail to be an equivalence relation for certain types of functions?
3. How do equivalence classes partition the set $X$?
4. How would the equivalence classes look if $f$ is injective?
5. How do the equivalence classes change if $f$ is a constant function?

Tip: When dealing with equivalence relations, remember that they naturally induce a partition of the set into disjoint equivalence classes.