The metric is the typical Euclidean metric, $ \sqrt{ (x_1 – x_2)^2 + (y_1 – y_2)^2 } $.
I have solved this one, albeit with in my opinion quite excessive steps. I would love to know if there is a simpler way to do it. Below is my approach.
Denote $A = \{(x_1,x_2) \in \mathbb{R}^2 : x_1 > x_2 \}$.
Take $p \in A$ and write $p = (x_1, x_2)$, where $x_1 > x_2$. Want to find a ball around $p$ entirely contained in $A$.
The smallest distance from $p$ to the line $x_2 = x_1$ is $\frac{x_1-x_2}{\sqrt 2}$.
Take any smaller radius, such as $ r = \frac{x_1 – x_2}{10} > 0$. Want to show the open ball $B_r (x) \subset A$.
Let $q \in B_r (x)$ and write $ q = (y_1, y_2)$. Want to show $y_1 > y_2$.
Since $q \in B_r (x)$, we have $\sqrt{ (x_1 – y_1)^2 + (x_2 – y_2)^2 } < r$.
This implies respectively
$$|x_1 – y_1| < r \ \ \ \text{ and } \ \ \ |x_2 – y_2| < r,$$
owing to $|a| \leqslant \sqrt{a^2 + b^2} < r$.
Adding these inequalities we get the condition $ |x_1 – y_1| + |x_2 – y_2| < 2r $.
Using the symmetry of absolute value, write $|x_1 – y_1| = |y_1 – x_1|$. The triangle inequality in reverse gives us:
$$|y_1 – x_1 + x_2 – y_2| \leqslant |x_1 – y_1| + |x_2 – y_2| < 2r$$
Exploiting the fact that $2r = \frac{x_1 – x_2}{5} < x_1 – x_2 $, we now write: $$ x_2 – x_1 < -2r < y_1 – x_1 + x_2 – y_2 < 2r < x_1 – x_2 $$
Adding and subtracting $x_1$ and $x_2$ respectively, we get:
$$ 0 < y_1 – y_2 < 2(x_1 – x_2) $$
Since $x_1 > x_2$, this inequality is legitimate and shows $y_1 > y_2$. Done.
Could there by a simpler way to show this? I found all the inequality juggling somewhat circuitous, which made me wonder whether there is a slightly more shorter and elegant way to show it.
Best Answer
You are off to a good start. Here is a shorter version of your argument: let $p \in A$ and write $p = (x_1,x_2)$ so of course $x_1 > x_2$. Define $r = x_1 - x_2$.
Select $0 < \epsilon < \dfrac r2$. If you are given a second point $q = (y_1,y_2)$ with $|p-q| < \epsilon$ then $$|x_1 - y_1| \le |p-q| < \epsilon$$ and $$|x_2 - y_2| \le |p-q| < \epsilon$$ so that $$y_2 \le x_2 + \epsilon = x_1 - r + \epsilon \le y_1 - r + 2\epsilon < y_1$$ and in particular, $q \in A$. It follows that $B(p,\epsilon) \subset A$.
This implies every $p \in A$ has a neighborhood contained in $A$ so that $A$ is open.