Show that $S = \{(a,a) | a \in A\}$ is an equivalence relation on A. How may equivalence classes does S define in A?

Question

For note $|A| = n\ge 1$

$aSa$ so S is reflexive.

I'm unsure how you prove symmetry and transitivity when there's only one element. For symmetry, would you say that $aSa$ has the reverse $aSa$? But transitivity still wouldn't work.

There are $n $ equivalence class, $[a] = {a}$, as there is only one element per $a $ for the whole set A.

Answer 1

It doesn't matter whether there is only one element or not. Your assumption only refers to one element at a time. Go back to the definition. Symmetry means that $(xSy) \implies (ySx)$ If $x \neq y$ then both clauses are false and the implication is true. If $x=y$ both clauses are true and the implication is true. Thus the relation is symmetric. The proof of transitivity is similar.

Answer 2

A relation $S$ is reflexive on $A$ exactly when $\forall x\in A~.xSx$.   Which is obviously the case when $S:=\{\langle a,a\rangle: a\in A\}$ .

A relation on $S$ is symmetric in $A$ exactly when $\forall x\in A~\forall y\in A~.(xSy\to yS x)$.   Well when $S=\{\langle a,a\rangle: a\in A\}$ we have that for any $x$ and $y$ in $A$: when $x=y$ then $xSy$ and $ySx$, else when $x\neq y$ then $\lnot(xSy)$ and $\lnot(ySx)$.   Therefore $xSy\to ySx$ is always valued as true, and so we have symmetry.

Likewise for transitivity. Show that there are no elements $x,y,z$ in $A$ which value $(xSy\land ySz\to xSz)$ as false.