Show that subset $T = \{(x,y) \mid x \neq 0, -3<y<3\}$ is open

I let $X=(x_1, y_1)$ be a point in $T$, let $B(r,X)$ be the ball of radius $r$ centered at $X$, and let $Y = (x_2,y_2)$ be a point in $B(r,X)$.

So i want to show that $Y$ is in $T$, in other words that $x_2$ not equal to $0$ and that $-3<y_2<3$.

I started with $\sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} < r$

...

$|y_2 - y_1| < r$ and $|x_2 - x_1| < r$

...

$y_1-r < y_2 < y_1+r$ and $x_1-r < x2 < x_1+r$

so I set $y_1 - r = -3$ and $y_1 + r = 3$

which led me assuming $r$ must be equal to the $\min(y_1 + 3, 3 - y_1)$

My questions are the following:

How do I use this to show that $Y$ is in $T$ ? What do I do with the fact that $X$ cannot equal $0$? Is my reasoning correct?

• Please note the edits, and check if they are correct or not. Also, pick up mathjax while you can, it is deadly simple and super useful on the site. Commented Sep 6, 2018 at 3:36

Your posting is a bit confusing. I assume you want construct an $$r \in \Bbb R$$ s.t. for all $$Y \in B(r,X)$$ also $$Y \in T$$ holds?

I started with $$\sqrt{(y_2-y_1)^2 + (x_2-x_1)^2} < r$$

...

$$|y_2−y_1| and $$|x_2−x_1|

If you mean an implication by "..." this is indeed true. So let's discuss what we actually want and what we have:

You ḱnow: $$X = (x_1,y_1) \in T$$ hence $$x_1 \not= 0, -3 < y_1 < 3$$

We want: $$Y=(x_2,y_2) \in T$$ hence $$x_2 \not= 0, -3 < y_2 < 3$$

So: How to choose $$r\in \Bbb R$$ s.t. the following implication holds:

$$|y_2−y_1|

1.) Let's ensure $$x_2 \not= 0$$ first. We know $$x_1 \not= 0$$ so if $$r < |x_1|$$ it follows directly $$x_2 \not = 0$$.

2.) To ensure $$-3 < y_2 < 3$$ we have choose $$r < 3 - y_1$$ as well as $$r < y_1 + 3$$ (this is what you already got as I realize now ^^)

So to ensure $$Y \in T$$ we have to choose $$r \in \Bbb R$$ this way it fulfills both 1.) and 2.)

So choosing $$r < \min\{|x_1|, y_1 + 3,3 - y_1\}$$ gives us the wanted result.

• This really helped thanks a lot. Commented Sep 7, 2018 at 13:07
• Let's explain more graphically what was the idea behind these formulas: We know $x_1 \not= 0$, so $x_1$ has a distance to $0$. The idea is now to ensure that the distance from $x_2$ to $x_1$ is less then the distance from $x_1$ to $0$. Then it follows directly that $x_2$ cannot be $0$. Now $|x_1|$ is the distance from $x_1$ to $0$ (because the absolute value can be seen as a distance). If we draw a circle around $x_1$ with radius $r < |x_1|$ all points within this circle cannot be $0$. And this circle is given by $$\left\{x_2 \Big| |x_1 - x_2| < r\right\}$$ That's it.
– Gono
Commented Sep 9, 2018 at 8:12