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I have to show that

$\mathbb{C}=\overline{\mathbb{C}\setminus\left\{0\right\}}$,

what is very probably an easy task; nevertheless I have some problems.

In words this means: $\mathbb{C}$ is the smallest closed superset of $\mathbb{C}\setminus\left\{0\right\}$.

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    $\begingroup$ Some problems? What problems? $\endgroup$ – Chris Eagle Jul 17 '13 at 14:31
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    $\begingroup$ What do you know about point-set topology? Do you know that it suffices to exhibit $0$ as a limit point of $\mathbb C\backslash \{0\}$? $\endgroup$ – Potato Jul 17 '13 at 14:33
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    $\begingroup$ This is a valid mathematical question that has a specific answer. Closing it as off-topic is just wrong. $\endgroup$ – Ayman Hourieh Jul 17 '13 at 15:18
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    $\begingroup$ I have no idea why this is getting closed. That's absurd. $\endgroup$ – Potato Jul 17 '13 at 16:58
  • $\begingroup$ If the topology of $\mathbb{C}$ is given by defining the Kuratowski closure operator, then this is true by definition. In a question like this, to get a precise proof, it might be good to also provide the definition of the topology you are using. $\endgroup$ – OR. Jul 17 '13 at 17:04
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The elements of the sequence of complex numbers $1$, $\frac{1}{2}$, $\frac{1}{3}$,$\cdots$ are all contained in $\overline{\mathbb{C} \setminus \{ 0 \} }$ but the limit of the sequence is $0$; hence $0$ is in the closure.

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  • $\begingroup$ I see but why is that enough why follows then that $\mathbb{C}$ is the closure? $\endgroup$ – math12 Jul 17 '13 at 14:45
  • $\begingroup$ The distance from $0$ to $\frac{1}{n}$ is $\frac{1}{n}$ and this distance goes to $0$ and $n$ goes to infinity. $\endgroup$ – Jay Jul 17 '13 at 14:48
  • $\begingroup$ Hm, I don't know if you got my problem or if I asked wrong... I know, that the closure consists of the limits of all convergent nets (especially all convergent sequences) with elements in $\mathbb{C}\setminus\left\{0\right\}$. Now, your sequence is in $\mathbb{C}\setminus\left\{0\right\}$ and converges to 0, so $0\in\overline{\mathbb{C}\setminus\left\{0\right\}}$. But why follows from this, that $\mathbb{C}=\overline{\mathbb{C}\setminus\left\{0\right\}}$. Thats not clear to me. Sorry. $\endgroup$ – math12 Jul 17 '13 at 14:55
  • $\begingroup$ The closure of a set contains all the limit points. $\endgroup$ – Jay Jul 17 '13 at 15:04
  • $\begingroup$ Hm, I guess I am blind then... i understand that it contains 0... but why $\mathbb{C}\setminus\left\{0\right\}$, too in order to get $\mathbb{C}$ as the closure?? $\endgroup$ – math12 Jul 17 '13 at 15:07
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Hint: (I assume you're using the usual topology on $\Bbb C$.) Note that $\{0\}$ is not an open set, so $\Bbb C\setminus\{0\}$ is not a closed set. $\Bbb C$ is closed, however. Is $\Bbb C$ a superset of $\Bbb C\setminus\{0\}$? If so, how many extra points does it have? What can you conclude?

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${\mathbb{C}\setminus\left\{0\right\}}$ is open, $\mathbb{C}$ is closed, and since $\overline{\mathbb{C}\setminus\left\{0\right\}}$ is the smallest closed set containing ${\mathbb{C}\setminus\left\{0\right\}}$, it must be equal to $\mathbb{C}$ (since the only set containing but not equal to ${\mathbb{C}\setminus\left\{0\right\}}$ is $\mathbb{C}$).

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$\partial(\mathbf{C}^\ast)$ consists of $0$ alone, use $\overline{\mathbf{C}^\ast}=\partial(\mathbf{C}^\ast)\cup\mathbf{C}^\ast$.

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