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Loser66
 one year ago
Which of the following subsets of C are open and which are closed?
1) {z : z<1}
2) the real axis
3)\(\{z:z^n =1~~for~~some~~integer~~n\geq 1\} \)
4)\(\{z\in \mathbb C: z ~~is~~real~~and~~0\leq z<1\}\)
5) \(\{z\in \mathbb C: z~~is~~real~~and~~0\leq z\leq 1\}\)
Please, help
Loser66
 one year ago
Which of the following subsets of C are open and which are closed? 1) {z : z<1} 2) the real axis 3)\(\{z:z^n =1~~for~~some~~integer~~n\geq 1\} \) 4)\(\{z\in \mathbb C: z ~~is~~real~~and~~0\leq z<1\}\) 5) \(\{z\in \mathbb C: z~~is~~real~~and~~0\leq z\leq 1\}\) Please, help

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It is easy to see that 1) is open 2) is open 3) is open 4) is not open nor closed 6) is closed But how to prove them briefly?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Not sure, not a set theorist. According to wikipedia, a set is closed if its complement is open. A subset \(U\) of the Euclidean nspace \(\mathbb{R}^n\) is called open if, given any point \(x\in U\), there exists a real number \(\epsilon > 0\) such that, given any point \(y\in \mathbb{R}^n\) whose Euclidean distance from x is smaller than \(\epsilon\), \(y\) also belongs to \(U\). Equivalently, a subset U of \(\mathbb{R}^n\) is open if every point in \(U\) has a neighborhood in \(\mathbb{R}^n\) contained in U.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1So I guess for 1: \[ S_1=\{z : z<1\}\\ \forall x\in S_1,xy<x1\implies y\in S_1 \]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0You define \(\varepsilon = x1\), right?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Yeah. I am really uncertain about this stuff though. I will be starting my university life (i.e. year 1) in around two weeks time. Ask someone else to check it lol.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0ok, thanks for being here. :)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1I am not so sure about the real line though. It doesn't seems open to me. No epsilon will work since you can only move in one direction to stay in real line.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0oh, it is closed. I am sorry. Since its complement is open. B = the real axis, it means the imaginary part =0 hence C  B = set of imaginary part is not 0. hence if let x in C  B , \(B(x, imaginary part/2)\) is belong to CB, hence B is closed

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.13 also doesn't seems to be open. Its complement should be open with \(\epsilon=\min\{x:k\in \mathbb{Z}^+\land 0\leq k<n,x=xz^n\}\).

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1What is B(x,imaginarypart/2)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The ball with center x and radius = imaginary part of x /2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if this ball belongs to the set, then the set is open

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.14 is closed since \(\epsilon=\min\{y:k\in \mathbb{R}\land 0\leq k < 1,y=xk\}\) works for its complement if we are allowed to construct such infinite set.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.15 is closed by a similar construction.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0just one small thing: the epsilon you define always /2 to have radius of the ball.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0because the point is a center. dw:1442000626403:dw

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Actually \[ S_1=\{z : z<1\}\\ \forall x\in S_1,xy<1x\implies y\in S_1\text{ not }x1 \] Not so sure about the divide by two part. Since \(\epsilon\) is the upper bound of the Euclidean distance, it should describe an open disk with radius \(\epsilon\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0have to divide by 2 because if you take x such that x = 0.8, then x is in S, right? but 1x = 0.2. hence if you take directly, that is the ball center x, radius 0.2, you get the bound of the circle. I meant the point ON the circle, which is not in S. If you divide by 2, pretty sure that the ball never touch the border. Hence it satisfies the definition of open set. dw:1442284284660:dw

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1But \(xy\color{red}{<}\epsilon\implies y\in U\).
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