Method of approach for Equivalence relations

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Let R be the binary relation on N×P(N) defined by (u, V )R(w, X) if and only if u×|V | $\leq$ w×|X|. Is R reflexive? Is R symmetric? Is R antisymmetric? Is R transitive? Is R a partial order?

I understand from this question that R is a powerset of natural numbers multipled by another set of natural numbers. But what i dont understand is how can it possibly be any of these, like you cant have a nat number set where 1 is equivalent to 2 and 2 is equivalent to 3,

or am i misunderstanding the question as a whole? I need help understanding the questions mainly.

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The set (call it) $A$ on which the relation has been defined is the cartesian product of two sets $\Bbb{N}$ and the power set of natural numbers $\mathcal{P}(\Bbb{N})$. $$A=\Bbb{N} \times \mathcal{P}(\Bbb{N})=\{(n,A)\, | \, n \in \Bbb{N} \text{ and } A \in \mathcal{P}(\Bbb{N})\}.$$ So examples of objects inside the set $A$ are the pairs $$\left(7, \{1,3,5,7,11,13\}\right), \quad (1, \emptyset), \quad (23, \{2^k \, | \, k \geq 0\})$$ where the first component is a natural number and the second component is a subset of $\Bbb{N}$.

Two elements $(n,A)$ and $(m,B)$ are related if $n\cdot |A| \leq m \cdot |B|$, where $|S|$ is the cardinality of set $S$.

For example, $(1, \emptyset)$ is related to $\left(7, \{1,3,5,7,11,13\}\right)$ because $0=1 \cdot 0 \leq 7 \cdot 6=42$.

Now we can explore the question in general.

Is the relation reflexive?

For that we need to determine if $(n,A)$ be related to itself for all $n \in \Bbb{N}$ and for all $A \in \mathcal{P}(\Bbb{N})$, i.e. can we guarantee that $$n \cdot |A| \leq n \cdot |A|\,?$$ Assuming that $0 \not\in \Bbb{N}$, this will be true for both finite and infinite subsets of $\Bbb{N}$. Hence reflexive.

Is the relation symmetric?

We ask if $(n, A)$ is related to $(m,B)$, should $(m,B)$ be related to $(n,A)$?

In the example I mentioned above, ask yourself, will $$\left(7, \{1,3,5,7,11,13\}\right) \text{ be related to } (1, \emptyset) \,?$$

Hopefully you can pick up from here.