If $A$ and $B$ are compact subsets of $\mathbb R$, then so is $\frac{A}B$. Given compact $A,B \subseteq \mathbb{R}$ with $\alpha := \inf(B) > 0$, define
$$\frac{A}B := \left\{\frac{x}y : x \in A 
\text{ and } y \in B\right\}.$$
Show that $\frac{A}{B}$ is compact.
I am thinking of using the following result given in a previous exercice but I don't know how to proceed.

$K \subset \mathbb R^n$ is compact if and only if for all $\{x_{n}\}$ such that $\{x_{n} : n\in \{1,2,\dotsc\}\}\subset K$, there exists a convergent subsequence $\{{x_{\phi(n)}}\}$ that converges to a limit $l \in K$.

When does the infimum come into place?
 A: The condition $\inf B>0$ is there only to assure that $0\notin B$ and that therefore $\frac AB$ makes sense.
Take a sequence $(a_n)_{n\in\mathbb N}$ of elements of $A$ and a sequence $(b_n)_{n\in\mathbb N}$ of elements of $B$. Then $\left(\frac{a_n}{b_n}\right)_{n\in\mathbb N}$ is a sequence of elements of $\frac AB$ and every sequence of elements of $\frac AB$ can be obtained by this process. The sequence $(a_n)_{n\in\mathbb N}$ has a subsequence $a_{n_k})_{k\in\mathbb N}$ which converges to some $a\in A$. And the sequence $b_{n_k})_{k\in\mathbb N}$ has a subsequence $(b_{m_k})_{k\in\mathbb N}$ which converges to some $b\in B$. So$$\lim_{k\to\infty}\frac{a_{m_k}}{b_{m_k}}=\frac ab\in\frac AB.$$
A: Do you know the results that the continuous image of a compact set is compact and that the product of two compact sets is compact. If so, then you can argue that $A \over B$ is the image of the compact subset $A \times B$ of $\Bbb{R}\times(\Bbb{R} \setminus \{0\})$ under the continuous function $(x, y) \mapsto {x \over y}$.
A: A subset of $\mathbb R$ is compact if and only if it is closed (i.e. it contains its limit points) and is bounded.
So $A$ is a bounded set. If $B $ would not be lower-bounded by the positive constant $\alpha$ (for example, if $1/n\in B$ for any $n$), then the set $A/B$ could not be bounded, because of the denominators tending to zero.
So: $A$ bounded and $B$ bounded from below by a positive constant imply that $A/B$ bounded. Moreover, $A$ and $B$ closed imply that $A/B$ is closed, so the set is compact. Note that it is not necessary that $B$ is upper bounded.
