TITLE: How is the relation "the smallest element is the same" reflexive?
QUESTION [11 upvotes]: Let $\mathcal{X}$ be the set of all nonempty subsets of the set $\{1,2,3,...,10\}$. Define the relation $\mathcal{R}$ on $\mathcal{X}$ by: $\forall A, B \in \mathcal{X}, A \mathcal{R} B$ iff the smallest element of $A$ is equal to the smallest element of $B$. For example, $\{1,2,3\} \mathcal{R} \{1,3,5,8\}$ because the smallest element of $\{1,2,3\}$ is $1$ which is also the smallest element of $\{1,3,5,8\}$.
Prove that $\mathcal{R}$ is an equivalence relation on $\mathcal{X}$.
From my understanding, the definition of reflexive is:
$$\mathcal{R} \text{ is reflexive iff } \forall x \in \mathcal{X}, x \mathcal{R} x$$
However, for this problem, you can have the relation with these two sets:
$\{1\}$ and $\{1,2\}$
Then wouldn't this not be reflexive since $2$ is not in the first set, but is in the second set?
I'm having trouble seeing how this is reflexive. Getting confused by the definition here.

REPLY [6 votes]: A binary relation $R$ over a set $\mathcal{X}$ is reflexive if every element of $\mathcal{X}$ is related to itself. The more formal definition has already been given by you, i.e. $$\mathcal{R} \text{ is reflexive iff } \forall x \in \mathcal{X}, x \mathcal{R} x$$
Note here that you've picked two different elements of the set to make your comparison when you should be comparing an element with itself. Also make sure you understand that an element may be related to other elements as well, reflexivity does not forbid that. It just says that every element must be related to itself.