# How to sketch $\{(x, y) \in \Bbb R^2 : -1 \leq \text{max}\{|x|,|y|\} \leq 1 \cup 2 \leq \text{max}\{|x|,|y|\} \leq 4\}$

After checking this answer, I did not quite get what I was trying to understand.

I have a routine exercise which states: Let $$f: \Bbb R^2 \rightarrow \Bbb R$$, $$(x,y) \rightarrow \text{max} \{|x|,|y|\}$$ Sketch $$f^{-1}([-1,1] \cup [2,4])$$

Ok, so here's how the answer goes for this:

By definition, $$f ^{-1}([-1, 1] \cup [2, 4]) = \{(x, y) \in \Bbb R^2 : -1 \leq \text{max}\{|x|,|y|\} \leq 1 \text{ or } 2 \leq \text{max}\{|x|,|y|\} \leq 4\}$$. Hence, an appropriate sketch is: But how do they sketch this? What are the steps for this?

Since change of $$x$$ to $$-x$$ and $$y$$ to $$-y$$ leaves the curves invariant, they symmetric in all quadrant. So consider only the first quadrant where $$x,y \ge 0$$.
(1) The first figure is bounded by four lines: $$x=0, y=0, x=1$$ and $$y=1$$. Finally it is a square with corners $$(1,1), (-1,1), (-1,-1),(1,-1)$$. So, it is a region with all points inside it and hence it is full grey.
(2) Similarly, $$max{|x|,|y|} \ge 2$$ indicates all points outside the square with corners $$(2,2),(-2,2),(-2,-2),(2,-2)$$ and $$max{|x|,|y|}\le 4$$ means all points inside the square with corners $$(4,4),{-4,4),(-4,-4),(4,-4)$$. So the required area is the overlap of these two squares and it is greyed..
• I have talked in terms of corners center as $(0,0)$ is understood.! Cheers. Mar 14, 2020 at 15:27