Determine whether the set

$$A=\{z\in\mathbb{C}\,\colon \Re{(z)}\ge0,\,\Im{(z)}\ge0,\, \vert z\vert\le1\}$$

is closed.

This is true if its complement is open. We have that

$$\overline{A} = \mathbb{C}\setminus\{z\in\mathbb{C}\,\colon \Re{(z)}\ge0,\,\Im{(z)}\ge0,\, \vert z\vert\le1\}$$

So I have split this complement into a union of finitely many complements:

$$\overline{A} = \bigcup_{i=1}^3 U_i$$


$$U_1=\{z\in\mathbb{C}\colon \Im{(z)}<0\}, \,\,U_2=\{z\in\mathbb{C}\colon\Re{(z)}<0\},\,\,U_3 = \{z\in\mathbb{C}\colon \vert z\vert >1\}$$

Now $U_1,U_2$ are clearly open (omitted detail for brevity). It is left to show $U_3$ is open. That is, if we take $z_0\in U_3$, then we must find an $r>0$ such that $D(z_0,r)\subseteq U_3$. Take

$$0<r<\vert z_0\vert -1\implies \vert z_0\vert-r>1.$$

Let $z\in D(z_0, r)$, then $\vert z_0-z\vert<r$. We want to show $\vert z\vert >1$. Now

\begin{align} r &> \vert z_0-z\vert \\ &\ge \vert z_0\vert - \vert z\vert \end{align}

$$\implies \vert z\vert > \vert z_0\vert - r>1$$

as required. Since this was proven for any arbitrary $z\in D(z_0,r)$ we have shown that $D(z_0,r)\subseteq U_3$. Since this was shown for an arbitrary $z_0\in U_5$, hence $U_5$ contains all its interior points and is therefore open. Note $\overline{A}$ is the union of three open sets which is therefore open. Thus $A$ is closed.

  • $\begingroup$ Your method is a bit overly complicated, but, more than that, you need to properly isolate the problem. You say "we must find an $r$...", and then say "we want to show $|z|>1$" without having found such an $r$. If you take some time to think about what $r$ might be (perhaps with a diagram), then the fact that $|z|>1$ should become obvious. $\endgroup$
    – Will R
    Commented May 7, 2016 at 19:30
  • $\begingroup$ @WillR I have modified my answer - does this look ok? $\endgroup$ Commented May 8, 2016 at 20:13
  • $\begingroup$ Well done, this is good. Better still would be to explicitly give such an $r$, e.g., choose $r=(|z_{0}|-1)/2$. Also, you switch from $U_{3}$ to $U_{5}$ at the end, but this is minor. $\endgroup$
    – Will R
    Commented May 8, 2016 at 20:48

3 Answers 3


You needlessly complicate the problem: denoting by $re(z) = \Re z, \ im(z) = \Im z, \ abs(z) = |z|$, notice that $A = B \cap C \cap D$ where $B = re^{-1} ([0, \infty)), \ C = im^{-1} ([0, \infty)), \ D = abs^{-1} ([0,1])$. Since the functions $re, im, abs$ are continuous and the intervals $[0,\infty)$ and $[0,1]$ are closed in $\Bbb R$, their respective preimages $B,C,D$ will also be closed in $\Bbb C$ (the preimage of a closed set through a continuous function is closed). Since the finite intersection of closed sets is closed, $A = B \cap C \cap D$ will also be closed.

  • $\begingroup$ Hi, sorry I have not covered these properties you have used. Could you prove this geometrically instead as this is what we have been taught. $\endgroup$ Commented May 7, 2016 at 20:27
  • 1
    $\begingroup$ @user2850514: Your decomposition of $\bar A$ into that union of 5 subsets is not correct. Using my notations, $\bar A = \bar B \cup \bar C \cup \bar D = \{ z \in \Bbb C : \Re z < 0\} \cup \{ z \in \Bbb C : \Im z < 0\} \cup \{ z \in \Bbb C : |z| > 1\}$. $\endgroup$
    – Alex M.
    Commented May 8, 2016 at 19:18

Since in the comments you asked for a geometrical proof, here we go:

First of all note that the set $A$ you defined is simply a 'quarter pie/unit ball' in the first quadrant of $\mathbb{R}^2$.

If you take an arbitrary point $z$ in the complemenet of $A$ you can define $r=d(A,z)=\inf_{z_A \in A}d(z_a,z)$, i.e. the shortest distance between $z$ and $A$ (note that here the structure of $A$ is crucial!). Now if you take an $\epsilon >0$ such that $ r> \epsilon >0$ you will obtain that the open ball $B_{\epsilon}=\{z'\in \mathbb{ C}\mid d(z',z) < \epsilon \}$ is completely contained in the complement of $A$.

So we conclude that $z$ is an inner point of $A^c$ and since $z$ was arbitrary we obtain that any $z$ in $A^c$ is an inner point and so $A^c$ is open.

This implies that $A$ is closed.


Using converging sequences seems the easiest way to prove it. $Re(z), Im(z)$ and the absolute value are all continuous functions and they all preserve the properties of the set $A$


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