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Let S=$\{z \in \mathbb{C}:z=a+bi, a>0;1>b; a,b\in \mathbb{Q}\}$

  1. Is S bounded?
  2. What are the limits points of S?
  3. Is S closed?
  4. What are the interior and boundary points?
  5. Is S open and connected?
  6. What is cl(S)?
  7. what is the complement of S?
  8. Is S compact?
  9. Is the closure of S compact?

Work thus far I assume from the way that S is defined that it can be considered a subset of $A=\{z \in \mathbb{C}:z=a+bi, a,b\in \mathbb{Q}\}$ and that A has many of the properties of $\mathbb{Q}$.

  1. S is not bounded as the magnitude of a is unrestricted?
  2. unsure
  3. S is not closed as any subset of $\mathbb{Q}$ is not closed.
  4. Again because S shares properties of $\mathbb{Q}$, $int(S)=\emptyset$ and every point in $S$ is a boundary point.
  5. The set is not open and none of the points are connected.
  6. $cl(S)=\{z \in \mathbb{C}:z=a+bi, a>0,b<1\}$
  7. unsure
  8. No because $S$ is neither closed or bounded?
  9. No because $cl(S)$ it is not bounded.
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    What is you definition of magnitude and of closed set? Why any subset of $\mathbb Q$ is not closed? This is false, because one point is a closed set. – 2017-01-29
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    @Wolfram Your questions answered in order. The absolute value of a number; A closed set is one that contains all of its limit points; Feel that a subsection of $\mathbb{R}$ that only includes only rational points would not be closed am I wrong. – 2017-01-29
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    Ah, I get what is the problem. Usually when someone writes $0$a$ and $b$ are between 0 and 1, not that $a>0,1 – 2017-01-29
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    I will edit the question to correct that – 2017-01-29
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    what happend to your answer $6)$? – 2017-01-29
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    @user190080 I assumed that the same area is part of the closure but irrational numbers are included as well. – 2017-01-29
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    So $a,b\in \mathbb R$? Anyway,would this be closed (remember the open intervalls)? – 2017-01-29

2 Answers 2

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Answers for modified question;

(1) as a set of $\mathbb{C}$, no.

(2) $[0,\infty]\times(-\infty,1]$.

(3) no.

(4) $S^o=\phi$

(5) No.

(6) $[0,\infty]\times(-\infty,1]$.

(7) $\mathbb{C}-S$

(8) No.

(9) yes.

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    what is hausdorff space? Could you also explain your answer to 2? – 2017-01-29
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    Hausdorff space is that every two points of it can isolate by isolated open balls (sets). – 2017-01-29
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    @AzJ do you know what the topological space means? Without that it is problematic to explain what does "Hausdorff" mean. However, I think this notion is off-topic here. – 2017-01-29
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  1. Correct.
  2. Any point of the kind $a+bi$ where $a\ge0, b\le1$ is the limit point. To see this, we should take such a point and for any $\varepsilon$ show that there is other point of $S$ on the distance less than $\varepsilon$. But we can for sure take $0b_1\in\mathbb Q$ very close to $b$ (also at the distance at most $\varepsilon/10$). The by definition of module in complex numbers we can deduce that $a_1+b_1i$ is a good point for us. Of course we also need to show that for any other point it is not a limit point, that is, in some neibourhood there are no points of $S$ - try to do it yourself.
  3. Now that we calculated the limit points we see that there are some of them not in $S$, e.g. $(0,0)$, thus $S$ is not closed.
  4. Your answer is correct, but saying that is because it shares properties of $\mathbb Q$ is vague and is not a rigorous proof. We can just find irrational point arbitrarily near to any point of $\mathbb C$ and so $S$ has empty interior. All found limit points are boundary by the same reasoning and all non-limits points are not boundary, because they do not have any point of $S$ in a small neighbourhood. (But generally non-limit point can still be boundary, iff it is an isolated point of the set.)