# Open set

1. Sep 23, 2013

### tuggler

1. The problem statement, all variables and given/known data

Show that the set of $$\mathbb{R}^2$$ given by $$S = \{(x, y) \in \mathbb{R}^2 : x > y\}$$ is open.

2. Relevant equations

3. The attempt at a solution

Why is S the half plane that has boundary given by the line $$y = -x?$$

2. Sep 23, 2013

### Staff: Mentor

It's not - the boundary is the line y = x. I suspect a typo.

3. Sep 23, 2013

### Zondrina

This is what I also thought when I read the problem.

S is open if and only if S=S°.

Last edited: Sep 23, 2013
4. Sep 23, 2013

### tuggler

Thanks for the clarification. I am still having trouble with this problem.

I know I must determine a radius r such that $$\{(x_1, y_2)\in \mathbb{R}^2: \sqrt{|x-y|^2 +|x_1-y_1|^2}<r\}$$

but how can I find such an r?

What I did was:

I located the region of the plane where $$x\gt y$$ then I place a point $$P=(x_1,y_1)$$ in it where it is fairly close to the line $$y=x$$.

Now I moved to the left from the point P until I hit the line y=x, which I marked Q then this Q has coordinates (y_1,y_1).

But I cant find an r. The answer given is $$\frac{x_1 - y_1}{\sqrt{2}}.$$

Which I don't know how they got?

5. Sep 24, 2013

### Staff: Mentor

That equation doesn't look right to me. Why is it (x1, y2), and why doesn't y2 show up on the right?