Let $K\subseteq\mathbb{C}$ be compact and non-empty. Show there exists a $z\in K$ such that $|z|=\inf\{|w|:w\in K\}$.

Question

Let $K\subseteq\mathbb{C}$ be compact and non-empty. Show there exists a $z\in K$ such that $|z|=\inf\{|w|:w\in K\}$.

Is such a $z$ in general unique? Give an example of a non-compact $K$ such that no such $z\in K$ exists.

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Oh boy, here we go. Since $K$ is non-empty there exists at least one $w\in K$ with $|w|\in \mathbb{R}$. Since $K$ is furthermore compact, there exists a lower bound in $\mathbb{R}$ for $|w|$ and hence by the infimum property of $\mathbb{R}$ necessarily an infimum. Now let $w=z$ and then $|w|=|z|$.

However, this $z$ is not in general unique, since the absolute value of complex numbers is defined as its distance from $0$, meaning there are infinitely many $w\neq z$ with $|w|=|z|$, e.g. $e^{i}=w$ and $e^{i\pi}=z$.

An example of a non-compact $K$ such that no such $z$ exists would be the set of all real numbers which is $\subseteq\mathbb{C}$.

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Is this argument sufficient or even correct?

Answer 1

There is a lower bound because the set $\{|w|\,|\,w\in K\}$ is non-empty and each of its elements is non-negative. Therefore, $0$ is a lower bound. You don't need compactness for this.

The sentence “Now let $w=z$ and then $|w|=|z|$.” makes no sense. What is $z$?

You can use the fact that the function$$\begin{array}{ccc}K&\longrightarrow&\mathbb R\\w&\mapsto&|w|\end{array}$$is continuous. Therefore, it has a minimum.

In order to prove that $z$ is, in general, not unique, provide an example. For instance, take $K=\{z\in\mathbb{C}\,|\,|z|=1\}$. Then any $z$ will do.

The real numbers don't provide a counter example, since $|0|$ is the minimum of the absolute function there. Take $\mathbb{C}\setminus\{0\}$, for instance.

Answer 2

The first paragraph of your solution is in error, or is at least incomplete.

The set $\{|w| : w \in K\}$ is a nonempty subset of $[0,\infty)$ and as you point out must have an infimum $I$. The problem is to show that there is a point $z \in K$ with $|z| = I$. You said to take $z=w$ without ever specifying $w$.

By definition of the infimum there exists, for each $n \ge 1$, a point $w_n \in K$ with the property that $|w_n| < I + \frac 1n$. In particular, $|w_n| \to I$.

Since $K$ is compact there exists a subsequence $\{w_{n_k}\}$ that converges to a point $z \in K$. Thus you have both $$|z - x_{n_k}| \to 0 \quad \text{ and} \quad |w_{n_k}| \to I.$$

Now apply the triangle inequality to find $$|z| \le |z - w_{n_k}| + |w_{n_k}| \to I.$$

Thus $|z| \le I$, but $|z| \ge I$ since $z \in K$. Thus $|z| = I$.