I have a question regarding graph sketching:
For example I have the following:
Let $S_1 = \{(x,y) : (x − 1)^2 + (y − 1)^2 < 1\}$ be a subset of the plane $P = \{(x,y) : x,y \in \mathbb{R}\}$.
Drawing this would be trivial since, it has the center of $(1,1)$ and radius of $r = 1$.
Now if I have something like $A = \{(x, y) : |x| \leq 1, |y − 1| \leq 1\}$.
We have two inequalities $ |x| \leq 1$ and $|y − 1| \leq 1$, when sketching the graph should I jsut draw the intersection between them (which leads to have small rectangle)? or I should I consider them as two separate sets (which result to have large areas). I'm not sure since the set takes two inequalities and no value would be repeated.
Approach one:
Approach Two:
The expression $|x|\leqslant1,|y-1|\leqslant1$ actually means $|x|\leqslant1\color{red}{\text{ and }}|y-1|\leqslant1$. So, it's the intersection of the sets $\{(x,y)\in\Bbb R^2\mid|x|\leqslant1\}$ and $\{(x,y)\in\Bbb R^2\mid|y-1|\leqslant1\}$, which turns out to be a square.
Besides, I don't see functions here.