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

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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

Mathematics
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It 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?
Not sure, not a set theorist. According to wikipedia, a set is closed if its complement is open. A subset \(U\) of the Euclidean n-space \(\mathbb{R}^n\) 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.

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So I guess for 1: \[ S_1=\{z : |z|<1\}\\ \forall x\in S_1,|x-y|<|x|-1\implies y\in S_1 \]
|dw:1441998460778:dw|
You define \(\varepsilon = |x|-1\), right?
Yeah. 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.
ok, thanks for being here. :)
I 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.
oh, 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 C-B, hence B is closed
3 also doesn't seems to be open. Its complement should be open with \(\epsilon=\min\{x:k\in \mathbb{Z}^+\land 0\leq k
What is B(x,imaginarypart/2)?
The ball with center x and radius = imaginary part of x /2
if this ball belongs to the set, then the set is open
|dw:1442000196119:dw|
4 is closed since \(\epsilon=\min\{y:k\in \mathbb{R}\land 0\leq k < 1,y=|x-k|\}\) works for its complement if we are allowed to construct such infinite set.
5 is closed by a similar construction.
Thank you so much. :)
just one small thing: the epsilon you define always /2 to have radius of the ball.
because the point is a center. |dw:1442000626403:dw|
Actually \[ S_1=\{z : |z|<1\}\\ \forall x\in S_1,|x-y|<1-|x|\implies y\in S_1\text{ not }|x|-1 \] 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 \(\epsilon\).
:)
have to divide by 2 because if you take x such that |x| = 0.8, then x is in S, right? but 1-|x| = 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|
But \(|x-y|\color{red}{<}\epsilon\implies y\in U\).

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