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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

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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 :)

fitzmerl duron
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Anthony
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  • It might be helpful to you to know use that the preimage under a continuous map of a closed (open) set is closed (open). For instance, 1(b) is closed for that reason. And, also, in a connected set, no set different from the whole and the empty set are clopen, so once you've got they're closed (open) you don't need to check the opposite. Finally, for compactness use the characterization of being closed and bounded. – Javi Feb 25 '18 at 22:00
  • You're reasoning in 1(b) is not correct. To use the convergence criterion, you must find a sequence which lies inside B but whose limit does not. – Javi Feb 25 '18 at 22:02
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    Just to elaborate on Javi's comment, the set of points where $cos(x)+e^{xy} \geq 1$ is closed since it is the preimage of $[1, \infty)$ under the continuous map $f(x,y)=cos(x)+e^{xy}$ (ask yourself why this is continuous). You can use similar reasoning for the others. – Exit path Feb 25 '18 at 22:03
  • @Javi, I don't really see how the convergence criterion for 1b isn't met by my explanation. Is it by what I said about the limit alternating between being in the set and not being in the set (as a result of the limit not existing)? – Anthony Feb 25 '18 at 22:09
  • @Anthony the sequence you chose wasn't inside B for all $n$ as you noticed, and that's a necessary condition to use that criterion. It's the limit which tells you whether the set is not closed. – Javi Feb 25 '18 at 22:11
  • Also, the sequence must converge to use it, because the closure is the set of limit points. – Javi Feb 25 '18 at 22:13
  • Makes sense. And could you clarify what exactly a preimage is? Seemed interesting – Anthony Feb 25 '18 at 22:14
  • @Anthony Maybe you know it as inverse image, because I can't believe you're doing topology without knowing that concept. Just in case, the preimage of a set $B$ under $f:A\to C\supseteq B$ is the set $f^{-1}(B)={x\in A: f(x)\in B}$. – Javi Feb 25 '18 at 22:17
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    @Javi wow yeah we've never discussed that in class. Our prof just kinda jumped into set classification using convergence and the ball theorem (I feel like there'd be more tricks up my sleeve if we had done stuff like that) – Anthony Feb 25 '18 at 22:23
  • Let us continue this discussion in chat. – Javi Feb 25 '18 at 22:25

1 Answers1

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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.

DanielWainfleet
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