Let $R$ be a relation on $\mathbb{N}\times \text{power set}(\mathbb{N})$, defined as $$(a,A)R(b,B)\iff A\setminus B\subseteq\{a,b\}$$
Determine which of the properties reflexivity, symmetry, antisymmetry, and transitivity the relation $R$ possesses. Determine whether it is a partial order or an equivalence relation.
It is reflexive.
I'm having trouble with symmetry.
It seems symmetric to me, though.
Let $(a, A) R (b, B)$. Therefore, $A\setminus B$ is a subset of $\{a, b\}$.
Is $(b, B) R (a, A)$ i.e., is $B\setminus A$ a subset of $\{b, a\} = \{a, b\}$?
Well, I don't know. So, I started constructing a counterexample.
We want $(a, A) R (b, B)$, i.e., $A\setminus B$ is a subset of $\{a, b\}$, BUT $(b, B)\require{cancel} \cancel{R}(a,A)$ , i.e., $B\setminus A$ is NOT a subset of $\{a, b\}$.
Does this mean that $B\setminus A$ is a proper superset of $\{a, b\}$ i.e., is the opposite of "subset" - "proper superset"?
Let's start constructing the differences: $A\setminus B= \{1, 2\}$ and let $a = 1, b = 2$. We want $B\setminus A$ to be a superset of $\{a, b\}$, so it must have at least one more element: $B\setminus A = \{1, 2, 3\}$. Well, is that possible? How should $A$ look like?
$A = \{1, 2, \ldots\}$ (due to $A\setminus B = \{1,2\}$). What about B? $B = \{3, 1, 2, \ldots\}$ (due to $B\setminus A = \{1, 2, 3\}$).
Well, this directly violates $A\setminus B =\{1,2\}$. This means that $B$ does not contain $2$ as an element. But then $B\setminus A$, on the other hand, contains $2$ as an element. It's impossible.
So, I think the relation $R$ is symmetric. If what I asked above is true:
Let's assume that $B\setminus A$ is a proper superset of $\{a, b\}$, i.e., $\{a, b\}$ is a proper subset of $B\setminus A$.
This means that $a, b$ are from the difference $B\setminus A$, i.e., $a, b$ are from $B$ but not from $A$.
We have $A\setminus B$ is a subset of $\{a, b\}$. From here, I want to conclude that $\{a, b\}$ is a subset of $A$, which directly contradicts "$a, b$ are not from $A$", but I'm not sure if I can.
Let's consider the general case $A\setminus B$ is a subset of $C$. Does this imply $C$ is a subset of $A$?
The standard procedure for proving "$C$ is a subset of $A$": let $c$ be from $C$. Well, yeah, but here we can't conclude anything from this.
We can take an element $x$ from $A$ but not from $B$, so it will also be from $C$. And what's the point of this?
Consider possible elements of $A$ and $B$ that are neither $a$ nor $b$ and how they impact the evaluation of the relations.
Let's say $c$ is such and consider all $A, B\subseteq\{a,b,c\}$.
Complete the truth table and examine the result.
$$\boxed{\begin{array}{c:c|c:c} c\in A& c\in B & A\smallsetminus B\subseteq\{a,b\}&B\smallsetminus A\subseteq\{a,b\}\\\hline \mathrm T&\mathrm T\\\hdashline\mathrm T&\mathrm F\\\hdashline \mathrm F&\mathrm T\\\hdashline \mathrm F& \mathrm F\end{array}}$$