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|<l\} $. 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 $x+y = 0$.
1
-
$\begingroup$ out of curiosity, is this from an undergraduate real analysis course? $\endgroup$– Hossien S 'MyMathYourMath'Jun 27 at 2:27
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Let $S = \{(x,y) \in \mathbb{R}^2: |k-x|+|m-y|<l\} $, 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 $B$.
-
$\begingroup$ question, isn't $(k,m)$ fixed, how do you know its $(0,0)$? $\endgroup$ Jun 27 at 2:15
-
3$\begingroup$ @HossienSahebjame we can shift the whole system. So, one fixed point is enough. $\endgroup$ Jun 27 at 2:20
-
$\begingroup$ I see, so its invariant under translations? $\endgroup$ Jun 27 at 2:21
-
1
-