# Open set, closed set or neither?

Is $P=\{(x,y)\in\mathbb R^2: x>0, y\geq0\}$ open, closed or neither? Proof your claim.

For me thix exercise is really difficult because we just reviewed the concepts in class and then got this exercise. I know how to proof if something is an open set, so i have to find an r that for all ϵ>0 $B_ϵ(x_0,y_0)$ s.t. $|(x,y)-(x_0,y_0)|<r$. If I now would try to proof that it is open and I get that it is not open, have I then to proof that is closed? And if its not closed I can follow that it is neither? My guess would be that is neither but I find it really hard to proof it.

• “Is it open?” and “is it closed?” are to separate questions. Just answer both, and you are done ;) Do you think that the set is open? That it is closed? What would you have to do to show that it is not open? That it is not closed? Commented Dec 4, 2014 at 12:01
• Does the point $(1,0)$ have a neighborhood contained in $P$? Does the point $(0,1)$ have a neighborhood contained in the complement of $P$? (Or the latter question for the point $(0,0)$ too?) Commented Dec 4, 2014 at 12:06

## 1 Answer

To prove that set $P$ is open (or not) you must check wether (or not) you can find for every $(x_0,y_0)\in P$ an $\epsilon>0$ such that the set $B_{\epsilon}(x_0,y_0):=\{(x,y)\in\mathbb R^2\mid ||(x,y)-(x_0,y_0)||<\epsilon\}$ is a subset of $P$.

Can you find a suitable $\epsilon$ for $(1,0)\in P$?

If not then the conclusion is that $P$ is not open.

Proving that $P$ is closed (or not closed) is the same as proving that $P^c$, i.e. the complement of $P$, is open (or not open).

Can you find a suitable $\epsilon$ for $(0,1)\in P^c$?

If not then the conclusion is that the complement $P^c$ is not open, or equivalently that $P$ is closed.