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I was trying to determine whether the following set is open or not:

$$C:=\{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z > 3 , z \ge -1 \} .$$

To do so, I tried to show that the following two other sets (whose intersection, $C_1 \cap C_2$, equals $C$) were both open: $$C_1:=\{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z > 3\}$$ $$C_2:=\{(x,y,z) \in \mathbb{R}^3 : z \ge -1\}$$

I found that $C_1$ was open, but that $C_2$ was not. If this is correct, I guess it means that $C$ is not open, but I was not 100% sure about that since the property I used only says that "the finite intersection of open sets is an open set". It doesn't say anything about the intersection of an open set with a closed set...

So, how can I be certain that $C$ is not open? And is there any way to show that such a set could be closed (or not) as well?

Thanks in advance for the help.

Travis Willse
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justdoit
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    "Usually" an intersection of an open and a closed set will be neither. There are exceptions, so you have to take each case on its own terms. – Angina Seng Apr 20 '17 at 14:12

1 Answers1

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All three outcomes are possible: For example, suppose $A$ open and $B$ closed:

  1. If $A \subset B$, then $A \cap B = A$ is open.
  2. If $B \subset A$, then $A \cap B = B$ is closed.
  3. In the case where the topological space is $\Bbb R$ endowed with the usual topology, $A = (-1, 1)$, and $B = [0, 2]$, the intersection is $A \cap B = [0, 1)$, which is neither open nor closed. For most topologies one encounters, this case is typical, at last when the intersection is nonempty.

This means we must generally handle each case separately.

Hint For the particular example, consider the points $(2, 0, -1)$ and $(3, 0, -1)$. Are these points in $C$? $\partial C$?

Travis Willse
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  • Thanks a lot for the clear answer. Considering the particular example, $(3,0,-1)$ is in $C$ and $(2,0,-1)$ is not, but I couldn't tell if they are in $\partial C$ or not (and even if I could, I don't really understand how this could help)... may I ask for some other hint or explanation so that things become a little bit more clear to me? – justdoit Apr 20 '17 at 14:48
  • Both are in $\partial C$, and one can see by considering the paths $\gamma_c(t) := (c, 0, t)$, $c = 2, 3$, that respectively pass through the given points. For example, $\gamma_2(t) \in C$ precisely for $t > -1$, so $\gamma_2(-1) = (2, 0, -1)$ is in $\partial C$ but not $C$. – Travis Willse Apr 20 '17 at 14:53
  • I actually began to study topology only 2-3 weeks ago and I didn't know what a path was (and I'm still not sure to know, because I just took a quick look on wikipedia, and they say that "a path in a topological space $X$ is a continuous function from the unit interval $[0,1]$ to $X$", but the paths you mention can take values such as $-1$...? I'm confused). Therefore I guess this is not the way I'm supposed to solve the exercise... However, I'm still curious and would like to understand what you try to explain me, so if you have more time to try... that would be welcome. :) – justdoit Apr 20 '17 at 15:11
  • What definition of open and closed are you using? – Travis Willse Apr 20 '17 at 15:15
  • For a metric space $(X,d)$, we wrote that a set $U \subset X$ is open if $\forall x \in U$, $\exists \epsilon > 0$ such that $B(x,\epsilon) \subset U$. We saw that a set $A \subset X$ is closed if and only if $X \setminus A$ is open. – justdoit Apr 20 '17 at 15:21
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    Okay, so in that case, consider the point $x_3 := (3, 0, -1) \in C$. For any $\epsilon > 0$, the ball $B(x_3, \epsilon)$ contains the point $(3, 0, -1 - \frac{1}{2}\epsilon)$, which is /not/ in $C$. Thus, $C$ is not open. On the other hand to show that $C$ is not closed, we must show that $\Bbb R^3 - C$ is not open. Can you adapt the argument for $x_3$ to show this by considering the point $(2, 0, -1)$? – Travis Willse Apr 20 '17 at 15:25
  • Thanks a lot! You're awesome. :) To show that $C$ is not closed: $x_2 := (2,0,-1)$ is in $\mathbb{R}^3 \setminus C := {(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z \le 3, z < -1}$, and the ball $B(x_2 , \epsilon)$ contains the point $(2+\frac{1}{2}\epsilon , 0, -1)$, which is not in $\mathbb{R}^3 \setminus C$. So $\mathbb{R}^3 \setminus C$ is not open, so $C$ is not closed. Is that it? – justdoit Apr 20 '17 at 15:41
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    Perfect! And you're welcome. – Travis Willse Apr 20 '17 at 15:44