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.
Best Answer
Why are you testing reflexivity by looking at two different elements of $\mathcal{X}$? The definition of reflexivity says that a relation is reflexive iff each element of $\mathcal X$ is in relation with itself.
To check whether $\mathcal R$ is reflexive, just take one element of $\mathcal X$, let's call it $x$. Then check whether $x$ is in relation with $x$. Because $x=x$, the smallest element of $x$ is equal to the smallest element of $x$. Thus, by definition of $\mathcal R$, $x$ is in relation with $x$. Now, prove that this is true for all $x \in \mathcal X$. Of course, this is true because $\min(x) = \min(x)$ is always true, which is intuitive. In other words, $x \mathcal{R} x$ for all $x \in \mathcal X$, which is exactly what you needed to prove that $\mathcal R$ is reflexive.
You must understand that the definition of reflexivity says nothing about whether different elements (say $x,y$, $x\neq y$) can be in the relation $\mathcal R$. The fact that $\{1\}\mathcal R \{1,2\}$ does not contradict the fact that $\{1,2\}\mathcal R \{1,2\}$ as well.