Determine whether the following subsets of $\mathbb{R}^2$ are open with respect to the metric $d$.

Question

Consider the plane $\mathbb{R}^2$ equipped with its standard metric $d$ given by

$$d((x_1,y_1),(x_2,y_2)) = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$$

Determine whether the following subsets of $\mathbb{R}^2$ are open with respect to the metric $d$.

Justify your answers.

a) $ A = \{(x,y) \in \mathbb{R}^2 | x \geq 0\}$.

I have been given the solution to this question but don't quite understand it. This is the solution I have been given:

Let $r > 0$.Denote the point $(0,0) \in \mathbb{R}^2$ by $0$ and consider the open ball $B_r(0) \subset \mathbb{R}^2$. For the point $p_r = (-\frac{r}{2}, 0)$ we have $$d(p_r,0) = \sqrt{\frac{r^2}{4}} = \frac{r}{2}<r$$

and therefore $p_r \in B_r(0)\subset \mathbb{R}^2$. Note that $0 \in A$. If $A$ were open, then we would be able to find a radius $r>0$ with the property that $B_r(0)$ is never entirely contained in $A$. Thus, $A$ is not an open set of $\mathbb{R}^2$ with respect to the metric $d$.

Looking at this solution I have a couple of questions. Why is the point $p_r = (-\frac{r}{2}, 0)$ used? I understand where the $d(p_r,0)$ comes from. If someone could help me break it down a bit. Thanks in advance.

Answer 1

The point of the proof is that it's not an open set because $(0,0)$ is not an interior point. For that matter $(0,k)$ for any $k$ is not an interior point, but the proof choose $(0,0)$ because it makes the math easier.

To show $(0,0)$ fails the interior test set we have to show there does NOT exist an $r > 0$ so that $B_r(0,0)$ is contained in $A$.

To do that we just have to find a point $(j,k)$ so that $d((0,0),(j,k)) < r$ and $(j,k) \in A$. If $(j,k)\not \in A$ then $j < 0$. And it doesn't matter what $k$ is. So to make the math easy we will let $k = 0$.

So we need a $(j,0)$ so that 1: $\sqrt{(j-0)^2 + (0-0)^2} = |j| < r$. and 2: $(j,0) \not \in A$ i.e. $j < 0$.

So we need $-r < j < 0$.

.... and they chose $\frac {-r}2$. It actually doesnt matter what you choose. They chose $(\frac {-r}2,0)$ because it makes the math easier.

The could have chosen $(-\frac{r}{e},10^{-57}r)$ but that would make our math hard as hell. [ $\sqrt{(-\frac{r}{e}-0)^2 + (10^{-57}r - 0)^2} < r$]

But with testing is $(\frac {-r}2,0)\in B_r(0,0)$ the math is easy.

Is $d((\frac {-r}2,0),(0,0)) < r$? Of course.

And is $(\frac {-r}2, 0) \in A$? Of course not.

So $A$ is not open.

Answer 2

There is a mistake at the end. If the set were to be open, then since $(0,0) \in A$, there must be a ball centered there entirely inside $A$, as opposed to what the solution says. Then you use the fact that for any $r>0$, $p_r$ is inside the ball of radius $r$ around the origin. Since this point is not in $A$, the ball can not be contained in $A$, so it can’t be open.