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Let $\mu\in \mathbb{C}$. Put $U=\{z\in \mathbb{C}, Re(z)>0\}$.

Now, we define de set $F_\mu=\{a\in U, \mu\in\{na,n\in N^* \} \}$.

I want to show that the set $F_\mu$ is closed in $\mathbb{C}$.

If $(a_p)\subset F_\mu$ such that $a_p\to a$.

Then for all $p\in N$, there exists $ n_{p}\in N$ such $\mu= n_{p}a_p$.

My question how i can to prove that the set is closed?

1 Answers 1

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Let $\mu=1$. Then $F_1=\{\frac{1}{n}: n \in \mathbb N\}$. $F_1$ is not closed.

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    But, why $F_1=\{\frac{1}{n}, n\in N^* \}$. – 2017-02-03
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    $a \in F_1 $ iff there is $n \in \mathbb N$ such that $1=na$ – 2017-02-03
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    Ok. Thank you very much – 2017-02-03
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    Other question. $F_\mu$ is closed on the complement of $U$. – 2017-02-03
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    $F_{\mu}$ is not a subset of the complement of $U$ ! We have $F_{\mu} \subseteq U$ ! – 2017-02-03
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    I want to say, is $F_\mu$ is closed set relatif of $U$ – 2017-02-03