Conceptualizing the (set theoretic) universe as a relation?

Question

The universe is the class $\mathcal{U} = \{x: x = x\} = \{x: x \text{ is a set}\}$.

Is there a way to define the universe in terms of a relation $R = \{(x,y): \psi(x,y)\}$, where $\psi(x,y)$ is some set theoretic formula in which $x$ and $y$ are the only free variables?

My question stems from having been given the following exercise:

What is $\mathcal{U}^{-1}$? Hint: recall that $R^{-1} = \{(x,y): (y,x) \in R\}$.

I am having difficulty making sense of the question itself as I am having a hard time conceptualizing the universe as a relation.

Edit: The notion of ordered pairs I am given is $(x,y) = \{\{x\},\{x,y\}\}$.

Answer 1

Assume the following definitions (as, for example, in Kelley's set theory) $$\{x\}=\{z:z\,\text{set}\implies z=x\}$$ $\{x,y\}=\{x\}\cup\{y\}$ and $(x,y)=\{\{x\},\{x,y\}\} $. Then $(x,y) $ is a set if and only if both $x $ and $y $ are sets. Consequently, we have \begin{align} \mathcal U^{-1} &=\{(y,x):(x,y)\in\mathcal U\}\\ &=\{(y,x):x,y\in\mathcal U\}\\ &=\mathcal U\times\mathcal U \end{align}

Answer 2

No, $\mathcal{U}$ is not a relation and there is not any useful way to conceptualize it as one. But the problem you are trying to solve in no way requires $\mathcal{U}$ to be a relation. Just use the definition: $$\mathcal{U}^{-1}=\{(x,y):(y,x)\in\mathcal{U}\}.$$ In other words, $\mathcal{U}^{-1}$ is the class of all ordered pairs $(x,y)$ such that $(y,x)\in\mathcal{U}$. For which $x$ and $y$ is $(y,x)\in\mathcal{U}$? If you can answer that, then you know what the elements of $\mathcal{U}^{-1}$ are.