I am trying to determine the interior and closure of the following set
$$ A := \{ (x,y) \in \mathbb{R}^2 : |x| \geq |y|\} $$
and would like to vertify my answer
For every $(x,y) \in A$ we can find an $\epsilon$-neighbourhood that is completely contained in $A$ unless $|x| = |y|$. If we let $a \in A$ such that $a = (x,x)$ and $\epsilon > 0$ then $ b = (x-\epsilon, x+\epsilon) \in B_{\epsilon}(a)$ and $ b\notin A$ which means $A^{°} := \{ (x,y) \in A : |x| \neq |y| \}$
The set $A$ is closed, which means $A = \overline{A}$. I tried to show that $\mathbb{R}^2 \setminus A$ is open, but I am having trouble finding an $\epsilon$ such that $$c \in \mathbb{R}^2 \setminus A \ \text{and} \ d = (x,y) \in B_{\epsilon}(c) \Longrightarrow |x| < |y| \ \text{and} \ d \in \mathbb{R}^2 \setminus A $$
Note: we are using the Euclidean metric on $\mathbb{R}^2$
Best Answer
Hint: try to draw the images of the set.
For all $(x_0,y_0)\in\mathbb R^2\setminus A$, $|x|<|y|$, thus $|x|\ne|y|$. Recall that the distance from a point $(x_0, y_0)$ to a straight line $Ax+By+C=0$ is $\frac {|Ax_0+By_0+C|} {\sqrt{A^2+B^2}}$. In this case, the distance from $(x_0, y_0)$ to $y=x$ and $y=-x$ are $\frac {|x_0-y_0|} {\sqrt 2}$ and $\frac {|x_0+y_0|} {\sqrt 2}$ respectively. From the fact $|x|\ne|y|$ we know that neither of the two distance can be zero. Let $r$ be the smaller one of these two, and the circle of radius $r/2$ is not in $A$. This can be verified immediately from the triangular inequality. That is the radius you have been looking for.