# Absolute Value Inequalities Question

Let $$(x,y) \in \mathbb{R}^2$$ and fix $$(k,m) \in \mathbb{R}^2$$ such that $$k+m=0$$. Let $$l>0$$ and consider the set $$S = \{(x,y) \in \mathbb{R}^2: |k-x|+|m-y|. How do I prove that for all $$l>0$$, $$S$$ will always contain some points of the set $$\{(x,y) \in \mathbb{R}^2: x+y<0\}$$ and some points of the set $$\{(x,y) \in \mathbb{R}^2: x+y>0\}$$. I know that (with the help of desmos) it is supposed to look like a square with a side length $$\sqrt{2}l$$ that lies symmetrically on the line $$x+y = 0$$.

• out of curiosity, is this from an undergraduate real analysis course? Jun 27 at 2:27

Let $$S = \{(x,y) \in \mathbb{R}^2: |k-x|+|m-y|, and $$A=\{(x,y) \in \mathbb{R}^2: x+y<0\}$$, and $$B=\{(x,y) \in \mathbb{R}^2: x+y>0\}$$.

For the sake of contradiction, assume that $$S \cap A =\phi$$.

Let $$p=(x,y) \in A$$, then $$x+ y < 0$$. Since $$(k,m)=(0,0) \in \mathbb{R}^2$$ then

$$|x+y|\leq |x-0|+|y-0| < l$$

Then $$p \in S$$. Therefore $$S \cap A \not =\phi$$.

Do the same for $$B$$.

• question, isn't $(k,m)$ fixed, how do you know its $(0,0)$? Jun 27 at 2:15
• @HossienSahebjame we can shift the whole system. So, one fixed point is enough. Jun 27 at 2:20
• I see, so its invariant under translations? Jun 27 at 2:21
• @HossienSahebjame yes. Jun 27 at 2:22
• makes sense now, thanks !! Jun 27 at 2:22