# Thread: Show that set is not open nor closed

1. ## Show that set is not open nor closed

I have a set I = {x from R2 : x1<1 v x2>=2}
I have to show it isn't neither open or closed.

Can I do this?:
x1<1then I can choose for that point an open disc of radius r = 1-x1
Every point y in that disc has 2x11<y1<1, so y1<1, so every point in the disk satisfies the first inequality so this is open

But if I do the same for x2, then I get 2<y1, so this isn't the same as x2>=2. Can I conclude that it isn't open? I can also say that if x2=2 you cannot say that there is an open ball in I with center z and radius r. So it isn't open.

What have I going to do?

2. ## Re: Show that set is not open nor closed

I would recommend reviewing the formal definitions of open and closed.

Consider the analogue case of [0,1) in R (with the usual topology on the reals). If you can show it's not open nor closed, then showing that set I is not open nor closed should be similar.

3. ## Re: Show that set is not open nor closed

The whole open/closed question bring to mind topology. I'm not sure what kind of set structures you are using in DG. There is a topology (the name escapees me at the moment) where [0, 1) is open. Or are we assuming we are using the "usual topology" on the reals?

-Dan

4. ## Re: Show that set is not open nor closed

Originally Posted by Cyn1
I have a set I = {x from R2 : x1<1 v x2>=2}
I have to show it isn't neither open or closed.
Can I do this?: x1<1then I can choose for that point an open disc of radius r = 1-x1
Every point y in that disc has 2x1−1<y1<1, so y1<1, so every point in the disk satisfies the first inequality so this is open.
The point $\mathcal{P}: (1,2)\notin\mathcal{I}$. Now if $\varepsilon >0$ then any ball $\mathscr{B}_{\varepsilon}(\mathcal{P})$ contains points in $\mathcal{I}$ and points not in $\mathcal{I}$, therefore $\mathcal{P}$ is a boundary point of $\mathcal{I}$ not in $\mathcal{I}$. That is enough to show that $\mathcal{I}$ is neither open nor closed.