Suppose $K \subseteq \mathbb{R}$ is compact and non-empty. Prove that $2K$ is compact

My Solution: Let $\{ O_{\lambda}\}_{\lambda \in \Delta}$ be an open cover of $2K$, where $\Delta$ is some index set. then $2K \subseteq \cup_{\lambda \in \Delta}O_{\lambda}$ and each $O_{\lambda}$ is open.\ Let $\frac{O_{\lambda}}{2}=\{\frac{x}{2}:x \in O_{\lambda} \}$, for each $\lambda \in \Delta$. Then each set $\frac{O_{\lambda}}{2}$ is open.\ \textit{Claim:} $K\subseteq \cup_{\lambda \in \Delta} \frac{O_{\lambda}}{2}$. To prove the claim, let $x\in K$. Then $2x\in 2K \subseteq \cap_{\lambda \in \Delta}O_{\lambda},$ so there exists $\lambda_{x}\in \Delta$ such that $2x\in O_{\lambda_{x}}$. Then $x=\frac{1}{2} 2x\in \frac{O_{\lambda_{x}}}{2}\subseteq \cup_{\lambda \in \Delta}\frac{O_{\lambda}}{2}.$ This establishes the claim. \ Therefore $\{\frac{O_{\lambda}}{2} :\lambda \in \Delta\}$ is an open cover of $K$. Since $K$ is compact, there exists some $n\in \mathbb{N}$ and $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n} \in \Delta$ such that $\{\frac{O_{\lambda_{1}}}{2},\frac{O_{\lambda_{2}}}{2},\frac{O_{\lambda_{3}}}{2}, \cdots, \frac{O_{\lambda_{n}}}{2} \}$ forms a finite subcover of $K$, which means that $K\subseteq \cup_{i=1}^{n} \frac{O_{\lambda}}{2}$.\ \textit{Claim:} $2K\subseteq \cup_{i=1}^{n}O_{\lambda_{i}}$. To prove the claim, let $y \in 2K$. So there exists $x\in K$ such that $y=2x$. Since $x\in K$ and $K\subseteq_{i=1}^{n}\frac{O_{\lambda_{i}}}{2}$, there exists some $j\in \{1,2,3,\cdots, n\}$ such that $x\in \frac{O_{\lambda_{j}}}{2}$. That is $x=\frac{z}{2}$ for some $z\in O_{\lambda_{j}}$. \ So $y=2x=2\frac{z}{2}=z\in O_{\lambda{j}}\subseteq \cup_{i=1}^{n}O_{\lambda_{i}}$. Thus, $2K$ is compact.

Solution 1:

As said in comments, you can also use :

• the fact that the image of a compact by a continuous map is still compact,
• the fact that compacts in $\mathbb{R}$ are exactly the closed and bounded subsets of $\mathbb{R}$.

Here is another approach, based on the sequential characterization of compactness : let $(y_n)_{n \in \mathbb{N}}$ be a sequence in $2K$. Then, for every $n$, one has $$\frac{1}{2}y_n \in K$$

so by compactness of $K$, the sequence $\left( \dfrac{1}{2}y_n \right)_{n \in \mathbb{N}}$ has a limit point $l$. You deduce immediately that $(y_n)_{n \in \mathbb{N}}$ has a subsequence converging to $2l$. This proves that every sequence in $2K$ has a converging subsequence, and hence, $2K$ is compact.