# Showing this set is closed in $\mathbb{C}$

Let $a \in \mathbb{C}$ and $r > 0$. Prove then that $B = \left\{x \in \mathbb{C} \mid | x- a | \leq r \right\}$ is closed.

Attempt: In class we defined a closed set as one whose complement is open. And a set $A \subset \mathbb{C}$ is open if $$\forall x \in A, \exists \delta > 0, \forall y \in \mathbb{C}: |y-x | < \delta \Rightarrow y \in A.$$

So I want to prove that $B^c = \left\{x \in \mathbb{C} \mid |x - a | > r \right\}$ is open. So let $x \in B^c$. Then $| x - a | > r$. I'm having trouble finding the correct $\delta > 0$. I see that $$r < | x - a | \leq | x - y | + | y - a|$$ I want to have $| y- a | > r$ so that $y \in B^c$. So how to pick $\delta > 0$ so that the term $|x - y |$ goes away in the above inequality?

• Try $\delta = |x-a|-r > 0$. – Henno Brandsma Jul 4 '16 at 11:13
• Try drawing a picture... – 5xum Jul 4 '16 at 11:22
• That doesn't seem to help me. Suppose then that $|x- y| < \delta = |x - a | - r$. Then I have $r < | x - y | + | y - a| < | x - a | - r + | y - a|.$ And then what? – Kamil Jul 4 '16 at 11:25

Let $(z_n)$ a sequence of $B$ that converge and let denote $\ell$ it's limits. Suppose $|\ell-a|>r$. Let $\varepsilon>0$ and $N$ s.t. $|z_N-\ell|<\varepsilon.$ Then $$r<|\ell-a|\leq|\ell-z_N|+|z_N-a|<\varepsilon+r$$
Since $\varepsilon>0$ is unspecified, we get $|\ell-a|=r$ which is a contradiction. Therefore, $\ell\in B$ and the claim follow.
• Why is $|z_N - a| < r$? – Kamil Jul 4 '16 at 11:32
• $|z_N-a|\color{red}{\leq} r$ because $z_N\in B$. – Surb Jul 4 '16 at 11:34
A set is closed iff it contains all of its limit points. Assume that $x \in B'$ and $|x - a| > r$. Let $|x-a| = r + \epsilon$ for some $\epsilon > 0$. The open ball $B(x, \epsilon/2)$ lays outside of $B$, since if $y$ is in this open ball, then by the triangle inequality we have
$$r + \epsilon = |x-a| \leq |a-y| + |y-x| < |y-a| + \epsilon/2$$
so that $|y-a| > r + \epsilon/2 > r$. Therefore, $x$ cannot be a limit point of $B$, since it has a neighborhood disjoint from $B$. This means that all limit points of the set lie inside the set, hence the set is closed.