# Proving an open subset of $\mathbb{R}^{2}$

I am having trouble finding a correct radius for a ball $B(\mathbb{x}, \epsilon)$ to prove the following statement:

Let $\Omega = \left\{ (x,y) \in \mathbb{R}^{2} : x + y \neq 0 \right\}$

Show that $\Omega$ is an open subset of $\mathbb{R}^{2}$.

So far I have tried to apply the triangle inequality:

$B(\mathbb{x}, \epsilon)$ = $\left\{\mathbb{y} \in \mathbb{R}^{2} : || \mathbb{x} - \mathbb{y} || < \epsilon \right\}$

$||\mathbb{x} - \mathbb{y} || \leq ||\mathbb{x}|| + ||\mathbb{y}||$

But do not know how to proceed or if this is even the correct direction for the proof.

• Draw a picture. – Paul Mar 11 '18 at 12:36

Hint

Method 1 : It may be easier to show that $$\Omega ^c=\{(x,y)\mid x+y=0\}$$ is closed.

Method 2 : It's easier to show that $\Omega$ is open for the norm $\|(x,y)\|_1:=|x|+|y|$ than for the norm $\|(x,y)\|_2:=\sqrt{x^2+y^2}$.

• So in particular for method 1, this involves showing that for sequence $\left\{x_k \right\}$ in the set that converges to $x_0$, that $x_0$ is also in the set? – Zac Sanchez Mar 11 '18 at 12:49
• yes, for example. You can also show that all point of $\Omega ^c$ is on the boundary of $\Omega ^c$. You can also show that $\Omega ^c$ is the graph of a continuous function... – Surb Mar 11 '18 at 12:52

If $x_n$ and $y_n$ are two convergent sequences with $x_n+y_n=0$, then (since $+$ is continuous) also their limits $x$ and $y$ satisfy $x+y=0$.

Hence the set $\{(x,y)\mid x+y=0\}$ is closed, thus the complement $\Omega$ is open.

Correct me if wrong:

Consider the line $L: y+x=0$.

Let $(a,b)$ be a point $\not \in \Omega$, i.e. not on $L$.

Distance: point $(a,b)$ to line $L:$

$d_L(a,b)=\dfrac{|a+ b|}{√2}>0.$

$d_L: \mathbb{R}×\mathbb{R} \rightarrow \mathbb{R}$ is continuous.

$d_L(a,b) >0$ :

There exists a $\delta > 0$, such that

$||(x,y)-(a,b)|| \lt \delta$ implies

$d(x,y) \gt 0$, i.e. $B_\delta (a,b) \subset \Omega$.

Hence $\Omega =${$(x,y)| x+y \not = 0$} is open.

Given a point $\vec{x} = (x,y) \in \Omega ;\ x\neq -y$. Choose $d= \frac{|x +y| }{4} \neq 0$. For all $(x_1, y_1) \in B_d(\vec{x})$ we get $|x_1 + y_1| =|x_1 - x + (x + y )+ y_1-y| \geq |x + y| - | y - y_1| - |x_1 - x | \geq \frac{|x + y|}{2} >0$, so $x_1 \neq -y_1$ as desired.

If $X=(x_0,y_0)$ is a point where $x_0+y_0 \ne 0$, the distance from $X$ to the line $x+y=0$ is found to be $$d= \frac {|x_0+y_0|}{\sqrt 2}$$

Thus a ball of radius $$\epsilon = d/2$$ and centered at X will be included in the region $x+y\ne 0$