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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

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  1. Loser66
    • one year ago
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    @thomas5267

  2. Loser66
    • one year ago
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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?

  3. thomas5267
    • one year ago
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    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.

  4. thomas5267
    • one year ago
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    So I guess for 1: \[ S_1=\{z : |z|<1\}\\ \forall x\in S_1,|x-y|<|x|-1\implies y\in S_1 \]

  5. Loser66
    • one year ago
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    |dw:1441998460778:dw|

  6. Loser66
    • one year ago
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    You define \(\varepsilon = |x|-1\), right?

  7. thomas5267
    • one year ago
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    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.

  8. Loser66
    • one year ago
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    ok, thanks for being here. :)

  9. thomas5267
    • one year ago
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    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.

  10. Loser66
    • one year ago
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    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

  11. thomas5267
    • one year ago
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    3 also doesn't seems to be open. Its complement should be open with \(\epsilon=\min\{x:k\in \mathbb{Z}^+\land 0\leq k<n,x=|x-z^n|\}\).

  12. thomas5267
    • one year ago
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    What is B(x,imaginarypart/2)?

  13. Loser66
    • one year ago
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    The ball with center x and radius = imaginary part of x /2

  14. Loser66
    • one year ago
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    if this ball belongs to the set, then the set is open

  15. Loser66
    • one year ago
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    |dw:1442000196119:dw|

  16. thomas5267
    • one year ago
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    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.

  17. thomas5267
    • one year ago
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    5 is closed by a similar construction.

  18. Loser66
    • one year ago
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    Thank you so much. :)

  19. Loser66
    • one year ago
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    just one small thing: the epsilon you define always /2 to have radius of the ball.

  20. Loser66
    • one year ago
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    because the point is a center. |dw:1442000626403:dw|

  21. thomas5267
    • one year ago
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    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 \(\epsilon\).

  22. Loser66
    • one year ago
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    :)

  23. Loser66
    • one year ago
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    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|

  24. thomas5267
    • one year ago
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    But \(|x-y|\color{red}{<}\epsilon\implies y\in U\).

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