Open set, closed set or neither?

Question

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.

Answer 1

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.

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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.