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Math Help - Complex open set proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Complex open set proof

    Q: Let U=\{z\in{\mathbb{C}}:0<|z|<1\}. Prove that U is an open set.

    A: Let z_{0}\in\\U by arbitrary. We need to find a r>0 such that the disk centered at z_{0}, D(z_{0},r) is completely contained in U. So, let r=min\{z_{0},1-z_{0}\}. Then for any z\in\\D(z_{0},r) we have that |z-z_{0}|<r. Therefore, z\in\\U and we are done.

    Is my choice of r legit? It seams reasonable that disk with r equal to the smaller of the two distance from the edge of the set will be in the set, since the neighborhood itself is open.

    Thanks
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  2. #2
    A Plied Mathematician
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    Well, I assume you meant r=\min\{|z_{0}|,1-|z_{0}|\}, since you can't compare complex numbers. Other than that, and assuming you're using open disks there, I think your general idea works.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Well, I assume you meant r=\min\{|z_{0}|,1-|z_{0}|\}, since you can't compare complex numbers. Other than that, and assuming you're using open disks there, I think your general idea works.
    Oh yeah, I forgot the vertical bars. Also, I am assuming that the disk is open.

    Thanks
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  4. #4
    A Plied Mathematician
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    Great. I think you've got it. You're welcome, and have a good one!
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  5. #5
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    Another way to do this using basic properties:

    Let U = \{z\in\mathbb{C}:0<|z|<1\}, \ \mathbb{S}^1 = \{z \in \mathbb{C}: |z|<1\}, \ \mathcal{O} = \{(0,0)\}

    Then U = \mathbb{S}^1 \setminus \mathcal{O} = \mathbb{S}^1 \cap \mathcal{O}^c, where A^c is the complement of A.
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