I'm doing some topology now and I'm honestly very confused on how to formally say that a set is open or closed or compact. Like especially with open sets, the proof for it (balls must exist of radius $r$) seems very arbitrary (since the radius can be as small as possible). It just makes more sense to me from a geometric standpoint (i.e drawing it out). Unfortunately, this doesn't suffice in my class so I'm asking for help for the following questions
My thoughts for the following questions are such:
1a) already solved
1b) Take the point $(n,0)$. At $n=0$, the point is in $B$, but if you take the point as $n$ approaches infinity, the point alternates between being both in and not in $B$, so by convergence, the set is not closed. Then I'm completely confused on how to do the ball proof to show that it is/is not open.
1c) $C$ is neither closed nor open. If you choose the point $(0,1,1/n)$, then it does lie in $C$. But if you take the limit as $n$ approaches infinity, then the point converges to $(0,1,1)$, which is not in the set. So it is not closed. Then if you place a ball on $(0,1,1)$, some points will lie outside of $C$, so it is not open.
1d) For starters, $D$ is a parabolic cylinder (I think). I have no idea what to do here.
I know a compact set is a set that is both bounded and closed.
2a) This set is bounded because of the squared terms and the equals sign. Since this set does contain its boundary points, this set must be closed? So it's compact?
2b) This set is an elliptic cone. I don't think it's bounded because it extends infinitely? And so it must not be closed either? Really confused here
2c) The left-hand side is unnecessary, since $e^x+e^y$ won't ever be zero or less than $0$. So if we take the point $(1/n,1/n)$, this exists in $C$, and the limit of the point as $n$ approaches infinity does also exist in $C$, so the set is closed. Then I think this set is also bounded (kind of more "duh" to me than any formal mathematical way to show it). So it's compact?
Any help or tips on how to approach these problems would be greatly appreciated :)
For Problem 1(b) let $f(x,y)=(\cos x) +e^{xy}.$ Observe that $p=(\pi /2,0)\in B$ because $f(\pi /2,0)=1.$
For any $r>0$ the open ball $B(p,r)$ of radius $r$ centered at $p$ contains the point $p'=(s+\pi /2,0),$ where $s=\min(r/2, \pi /4).$
And $p'\not \in B$ because $f(s+\pi /2, 0)=(\cos (s+\pi /2))+1=(-\sin s)+1<1.$ So $B $ is not open.