Showing two sets defined with the set builder notation are equal

Question

I have recently got into set theory, and just recently started dealing with proofs involving set inclusions. I tried tackling some problems and was successful in doing so, up until I got to the following question, which I, unfortunately, got stuck on:

I know that to show that both sets are equal, it's required to show that both $S⊆T$ and $T⊆S$, but I don't really know how to show it. I have tried to brainstorm and thought about solving for x in the first set, but it didn't help me much, and I got stuck.

I am not exactly sure what am I missing...

Could anybody please help me get on track?

Answer 1

$S \subseteq T$: Let $(x,y) \in S$. Then $(2-x)(2-y) < 2(4-x-y)$, $4-2x-2y+xy < 8-2x-2y$, $xy < 4$. Since $x,y \in \mathbb{N}$, $y \geq 1$ so $x < 4$ and by swapping $x$ and $y$, $y < 4$. Checking all nine cases for $x,y\in[1,3]$, we find $(x,y) \in T$.

You could go from $x <4$ to check three cases: $x = 1 \implies y \leq 3$, $x = 2 \implies y = 1$ and $x = 3 \implies y = 1$, also yielding $(x,y) \in T$.

Answer 2

Note: you can use algebra to show that $S=\{\langle x,y\rangle\in \Bbb N^2:xy < 4\}$

So you need to show that $\forall \langle x,y\rangle\in\Bbb N^2~.(xy<4\leftrightarrow \langle x,y\rangle\in T)$

For instance, $1y<4\leftrightarrow y\in\{1,2,3\}$. And so forth.