What is difference between $\{(x,y): x< 2 \implies y >2\}$ and $\{(x,y): y > 2 \implies x <2\}$

Question

What is difference between $\{(x,y): x< 2 \implies y >2\}$ and $\{(x,y): y > 2 \implies x <2\}$?

Let $S_1 = \{(x,y): x< 2 \implies y >2\}$, and $S_2 = \{(x,y): y > 2 \implies x <2\}$. So $(1,3), (1,4),$ and $(1.5, 2.1)$, they are all in $S_1$ and in my opinion, it seem like they are also in $S_2$. I know that the set conditions in $S_1$ and $S_2$ are not equivalent. Also my professor said $S_1$ and $S_2$ are not the same sets. If I try to write $S_2$ in this way: $S_2 = \{(x,y): y \le 2 \implies x \ge 2\}$, will that work?

Answer 1

As others have stated, you must remember that an implication is true if the antecedant and consequent are both true, xor the antecedant is false.   You've examined only elements that were in the former (the intersection), neglecting those in the later (the difference).

$\{(x,y): x<2\to y>2\} = \{(x,y): x<2\wedge y>2\}\cup\{(x,y): x\geq 2\}\\ \{(x,y): y>2\to x<2\} = \{(x,y): x<2\wedge y>2\}\cup\underbrace{\{(x,y): y\leq 2\}}$

Answer 2

By definition, we have

$$S_1=$$ $$[2,+\infty)\times\Bbb R\cup (-\infty,2)\times(2,+\infty) $$ and

$$S_2=$$ $$\Bbb R\times(-\infty,2]\cup(-\infty,2)\times(2,+\infty) .$$

it is clear that $S_1\ne S_2$.

Answer 3

Just one of an infinite number of examples:

Answer 4

Letting $P(x)$ be $x<2$, and $Q(y)$ being $y>2$, we have that

$$S_1=\{(x,y):P(x)\implies Q(y)\}\ \text{and}\ S_2=\{(x,y):Q(y)\implies P(x)\}.$$ Note that the condition required of $(x,y)$ to be in $S_2$ is the converse of the condition required by $S_1$. So to show $S_1$ and $S_2$ are different, construct a truth table with $P$, $Q$, $P\implies Q$, $Q\implies P$, and identify the rows where the implications have different truth values. Then for any of the rows, find some $(x,y)$ such that $P$ and $Q$ have the truth values given in that row.

Answer 5

Note that $\,\big(x \lt 2 \implies y \gt 2\big)$ $\iff \big(\lnot(x \lt 2) \lor (y \gt 2)\big)\,$, so $\,S_1=\{(x,y) : x \ge 2 \lor y \gt 2\}\,$. Similarly, $\,S_2 = \{(x,y) : x \lt 2 \lor y \le 2\}\,$.

It then becomes clear that, for example, $\,S_1 \setminus S_2 \ne \emptyset\,$, so the two sets are not identical:

$$ \begin{align} S_1 \setminus S_2 &= \{(x,y) : \big(x \ge 2 \lor y \gt 2\big) \land \lnot\big(x \lt 2 \lor y \le 2\big)\} \\ &=\{(x,y) : \big(x \ge 2 \lor y \gt 2\big) \land (x\ge2)\land(y \gt 2)\} \\ &=\{(x,y):(x\ge2)\land(y \gt 2)\} \end{align} $$