Closed in $\mathbb{R}^n$

Question

Let $A$ be a compact set of $\mathbb{R}\setminus\{0\}$ and $B$ be a closed subset of $\mathbb{R}^n$. Prove that the set $\{a \cdot b \vert a \in A, b\in B\}$ is closed in $\mathbb{R}^n$

Answer 1

Let's call this final set $C$. Take a sequence $c_{n}$ in $C$ which converges to some $c \in \mathbb{R^{n}}$. We want to show $c \in C$. Well for every $n$, $c_{n}=a_{n}b_{n}$ for some $a_{n} \in A$, $b_{n} \in B$. $a_{n}$ is a sequence in a sequentially compact space, so it has a convergent subsequence $a_{n_{k}} \to a \in A$. $A$ is closed in $\mathbb{R^{n}}\setminus \{0\}$, so $a \ne 0$. A subsequence of a convergent sequence converges to the same limit, so $c_{n_{k}} \to c$. Then, by standard theorems about limits,
$$b_{n_{k}}=\frac{c_{n_{k}}}{a_{n_{k}}} \to \frac{c}{a}$$ But this limit is in $B$ because $B$ is closed. So $c=a \cdot \frac{c}{a} \in C$ as required.

Answer 2

Product of compact and closed in topological group is closed

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