1. ## Open sets

I'm quite new to topology. Just wondering if anyone can help me out a bit.

Question
Let $\displaystyle \Omega = \{(x,y) \in \mathbb {R}^{2}: x + y \neq 0 \}$ Show $\displaystyle \Omega$ is an open subset of $\displaystyle \mathbb {R}^{2}$

My Attempt
Let $\displaystyle \bold {x}_{o} = (x_1,x_2) \in \Omega$.

We want to show that there exists an $\displaystyle \varepsilon > 0$ such that $\displaystyle B(\bold {x}_{o}, \varepsilon ) \subset \Omega$.

Suppose that $\displaystyle \bold {a} = (a_1,a_2) \in B(\bold {x}_{o}, \varepsilon )$.
Then, $\displaystyle d_\infty(\bold {a},\bold {x}_{0})< \varepsilon$.
So, $\displaystyle max \{ |a_1 - x_1|, |a_2 - x_2| \} <\varepsilon$.
$\displaystyle \Rightarrow |a_1 - x_1| < \varepsilon \ \ or \ \ |a_2 - x_2|< \varepsilon$
$\displaystyle \Rightarrow -\varepsilon + x_1 < a_1 < \varepsilon + x_1 \ \ or \ \ -\varepsilon + x_2 < a_2 < \varepsilon + x_2$
Thus, if we add those inequalities, we have $\displaystyle -2\varepsilon +x_1 + x_2 <a_1+a_2 \ \ or \ \ a_1+a_2 < 2\varepsilon +x_1 + x_2$

Then, choose $\displaystyle \varepsilon = min \{\frac{1}{2}(x_1 + x_2),-\frac{1}{2}(x_1 + x_2)\}$ for $\displaystyle a_1 + a_2 < 0 \ \ or \ \ a_1 + a_2 > 0$.

Thus, $\displaystyle \bold {a} \in \Omega$ and $\displaystyle d_\infty(\bold {a},\bold {x}_{0}) \subset \Omega$

Hence, $\displaystyle \Omega$ must be open.

Comment
Is the above working correct?

Is using the infinity distance function $\displaystyle d_\infty(\bold {a},\bold {x}_{0})$ instead of the euclidean distance function, mathematically correct? Could I have used something like $\displaystyle d_1(\bold {a},\bold {x}_{0})$ instead?

Is there anyway to simplify the working out? What is the best way choosing an epsilon?

2. Originally Posted by shinn
I'm quite new to topology. Just wondering if anyone can help me out a bit.

Question
Let $\displaystyle \Omega = \{(x,y) \in \mathbb {R}^{2}: x + y \neq 0 \}$ Show $\displaystyle \Omega$ is an open subset of $\displaystyle \mathbb {R}^{2}$

My Attempt
Let $\displaystyle \bold {x}_{o} = (x_1,x_2) \in \Omega$.

We want to show that there exists an $\displaystyle \varepsilon > 0$ such that $\displaystyle B(\bold {x}_{o}, \varepsilon ) \subset \Omega$.

Suppose that $\displaystyle \bold {a} = (a_1,a_2) \in B(\bold {x}_{o}, \varepsilon )$.
No! you can't talk about such a B until you have shown that it exists!

Better: if $\displaystyle (x_1, y_1)\in \Omega$ then it is NOT on the line x+ y= 0. Therefore, the distance from $\displaystyle (x_1, y_1)$ to that line is not 0. Let $\displaystyle \epsilon$ be that distance and show that no point in the neighborhood of $\displaystyle (x_1, y_1)$ with radius $\displaystyle \epsilon$ is on that line.

Then, $\displaystyle d_\infty(\bold {a},\bold {x}_{0})< \varepsilon$.
So, $\displaystyle max \{ |a_1 - x_1|, |a_2 - x_2| \} <\varepsilon$.
$\displaystyle \Rightarrow |a_1 - x_1| < \varepsilon \ \ or \ \ |a_2 - x_2|< \varepsilon$
$\displaystyle \Rightarrow -\varepsilon + x_1 < a_1 < \varepsilon + x_1 \ \ or \ \ -\varepsilon + x_2 < a_2 < \varepsilon + x_2$
Thus, if we add those inequalities, we have $\displaystyle -2\varepsilon +x_1 + x_2 <a_1+a_2 \ \ or \ \ a_1+a_2 < 2\varepsilon +x_1 + x_2$

Then, choose $\displaystyle \varepsilon = min \{\frac{1}{2}(x_1 + x_2),-\frac{1}{2}(x_1 + x_2)\}$ for $\displaystyle a_1 + a_2 < 0 \ \ or \ \ a_1 + a_2 > 0$.

Thus, $\displaystyle \bold {a} \in \Omega$ and $\displaystyle d_\infty(\bold {a},\bold {x}_{0}) \subset \Omega$

Hence, $\displaystyle \Omega$ must be open.

Comment
Is the above working correct?

Is using the infinity distance function $\displaystyle d_\infty(\bold {a},\bold {x}_{0})$ instead of the euclidean distance function, mathematically correct? Could I have used something like $\displaystyle d_1(\bold {a},\bold {x}_{0})$ instead?

Is there anyway to simplify the working out? What is the best way choosing an epsilon?