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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?

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  • $\begingroup$ Try $\delta = |x-a|-r > 0$. $\endgroup$ – Henno Brandsma Jul 4 '16 at 11:13
  • $\begingroup$ Try drawing a picture... $\endgroup$ – 5xum Jul 4 '16 at 11:22
  • $\begingroup$ 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? $\endgroup$ – Kamil Jul 4 '16 at 11:25
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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.

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  • $\begingroup$ Why is $|z_N - a| < r$? $\endgroup$ – Kamil Jul 4 '16 at 11:32
  • $\begingroup$ $|z_N-a|\color{red}{\leq} r$ because $z_N\in B$. $\endgroup$ – Surb Jul 4 '16 at 11:34
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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.

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