More about compact.

Question

Any advice to solve the following problem?

Let $K$ be a compact set of the real numbers. Prove that the set $|K|=\{|x|: x\in K\}$ is a compact set.

Thanks a lot!

Answer 1

Solution 1:

Since $K$ is compact, it is bounded (above and below), so it is clear that $|K|$ is also bounded. It remains only to show that $|K|$ is closed: Let $(x_n)\subseteq K$ such that $|x_n|\rightarrow y$, so in particular $y\geq 0$. We need to show that $y\in |K|$.

Notice that there are either infinitely many numbers $n$ such that $x_n\leq 0$ or such that $x_n\geq 0$, so we can take a subsequence $(x_{n_k})$ consisting of only positive or negative numbers. We have two cases:

Case 1: $x_{n_k}\geq 0$

Then $|x_{n_k}-y|=||x_{n_k}|-y|\rightarrow 0$, so $x_{n_k}\rightarrow y$. Since $K$ is compact, hence closed, then $y\in K$, so $y=|y|\in K$.

Case 2: $x_{n_k}\leq 0$

Then $|x_{n_k}-(-y)|=|(-x_{n_k})-y|=||x_{n_k}|-y|\rightarrow 0$, so $x_{n_k}\rightarrow -y$. Again, $-y\in K$, so $y=|-y|\in |K|$.

Solution 2:

First, show that $-K=\left\{-x:x\in K\right\}$ is compact (this should be easy enough). Then $-K\cup K$ is closed and bounded, so it is compact. Finally, $|K|=(K\cup -K)\cap [0,\infty)$, so it is again closed and bounded, hence compact.

Answer 2

Let $$K_+=\{x\in K\mid x\ge 0\},\qquad K_-=\{x\in K\mid x\le 0\}$$ They are both closed and bounded since $[0,\infty),(-\infty,0]$, and $K$ are closed, and $K$ is bounded. Now $$|K|=K_+\cup (-K_-)$$ Can you show that $-K_-=\{-x\mid x\in K_-\}$ is closed?