Let $X$ be the set of all nonempty subsets of $\{1, 2, 3\}$. Then
$X= \{\{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$
Define a relation $ R $ on $X$ as follows: For all $A$ and $B$ in $X$,
$A \, R\, B\Leftrightarrow$ the least element of $A$ equals the least element of $B$ .
Prove that $R$ is an equivalence relation on $X$.
The solution is:
$R$ is reflexive: Suppose $A$ is a nonempty subset of $\{1, 2, 3\}$. [We must show that $A \, R \, A,$ It is true to say that the least element of $ A$ equals the least element of $A$. Thus, by definition of $R$, $A \, R \, A$. (There is more, and the answers are proving the Symmetric and Transitive...But I just want to know the examples. I can connect the dots once I find an example based on proving the reflexive property of equivalence relations on this set).
My question is, what is this least element they are speaking of? Can someone give the least element based on the subsets. Thank you so very much.
You are over-doubting the assignment. The ordering is just the one of natural numbers restricted to the subset $\{1,2,3\}$, i.e. $1<2<3$. The "least element of a subset $S$" (allow me to call it $\min S$) is a proper element of $S$, not a singleton. Then, for instance $$\min\{1,2\}=1,\ \min\{2\}=2,\ \min\{1,2,3\}=1$$
Insight on the problem: To simplify the proof, you may notice that $$A \ R\ B\iff \min A=\min B$$ $\min$ is a function assigning to each $A\in\mathcal P(\{1,2,3\})\setminus\{\emptyset\}$ a value (in this case, an element of $\{1,2,3\}$). It is a general fact that, if $\mathcal R$ is a relation on $X$ and there exists a function $f:X\to Y$ such that $x_1\mathcal R x_2\iff f(x_1)=f(x_2)$, then $\mathcal R$ is an equivalence relation. The converse holds as well: if $\mathcal R$ is an equivalence relation on a set $X$, there exists a set $Y$ and a function $f:X\to Y$ such that $x_1\mathcal R x_2\iff f(x_1)=f(x_2)$. In general, though, $Y$ is not required to be a subset of $X$.