Proving open set without Cauchy-Schwarz or Preimage

Question

Edit: Please prove this without using Cauchy-Schwarz or Preimage. (Also, please show how a chosen $ \epsilon $ would work by proving that $B(\mathbf x, \epsilon)\subseteq\Omega $)

I've been having trouble proving the following set is open without the use of Cauchy-Schwarz inequality.

$$ \Omega = \{(x,y)\in \mathbb R^2:x+y\ne0 \} $$

Our definition of open is:

A set $S $ is open $ \iff $ every point is an interior point (i.e. $ \forall \mathbf x\in S $ there exists $\epsilon >0 $ such that $ B(\mathbf x,\epsilon)\subseteq S) $

I have seen a proof for this using Cauchy-Schwarz inequality however I am not very familiar with it so I would appreciate if someone could share a simpler way to prove this statement. (I am familiar with the regular triangle inequality $ ||\mathbf a+\mathbf b|| \le ||\mathbf a|| + \mathbf||\mathbf b || $)

I'm also not very familiar with much of the terminology associated with metric spaces since I am only doing an introductory course to Analysis at the moment, so apologies if any part of this description is ambiguous or unclear.

My Attempt

Suppose $\mathbf x =(x,y)\in \Omega$. We need to prove it is an interior point.

Geometrically, it is clear that taking the radius for the ball around $ \mathbf x $ as $ \epsilon=\frac{|x+y |}{\sqrt{2}} $ will work. (The perpendicular distance from the point to the line $ x+y=0 $).

However to prove this analytically (in case I was working on a similar question where I did not have geometric intuition to help me), I would like to show that

$$ B\bigg(\mathbf x, \frac{|x+y |}{\sqrt{2}}\bigg) \subseteq\Omega $$

So I need to show $ \forall \mathbf{a}\in B(\mathbf x,\epsilon), \mathbf{a}\in \Omega $.

I'm stuck at this point. My only guess would be to incorporate $$ ||\mathbf x - \mathbf a || < \epsilon $$ at some point but honestly I have no idea.

Answer 1

You can make your life easier by choosing a smaller $\varepsilon$. Let’s take $0<\varepsilon<\dfrac{|x+y|}{2}$. Now assume $\|(x,y)-(x’,y’)\|<\varepsilon$. Then $|x-x’|<\varepsilon$ and $|y-y’|<\varepsilon$. Now \begin{align} |x’+y’|&\ge|x+y|-|(x+y)-(x’+y’)|\\ &\ge|x+y|-|x-x’|-|y-y’|\\ &\ge|x+y|-2\varepsilon>0. \end{align} The first two inequalities follow from the triangle inequality.

Answer 2

Why Cauchy schwartz should be used here ? (Btw, I don't really see how it can be used...). Anyway,

Method 1: $$\Omega ^c=\{(x,-x)\in\mathbb R^2\mid x\in \mathbb R\},$$

which is obviously sequentially closed, and thus closed.

Method 2:

The function $f(x,y)=x+y$ is "obviously" continuous on $\mathbb R^2$. Moreover, $$\Omega =f^{-1}\big(\mathbb R\setminus \{0\}\big),$$ and thus $\Omega $ is open.