Prove that the given set is closed.

Question

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

Let $S=\{a·b | a ∈ A,b ∈ B\}$. We know that $S \subset \overline{S}$. For the converse, Let $y\in \overline{S}$, then $\exists \{y_n\}\subset S$: $\lim_{n\to \infty}y_n=y$. Let $y_n=a_n.b_n, a_n\in A$ and $b_n\in B$. $A$ is compact $\implies \exists \{a_{n_k}\}$ subsequence converges to $a_0$(say). $$b_{n_k}=\frac{1}{a_{n_k}}y_{n_k}.$$ Hence, $b_{n_k}$ converges to $b_0\in B$(say). subsequence $\{a_{n_k}.b_{n_k}\}$ converges to $a_0.b_0$. Hence, $\{a_n.b_n\}$ also converges to $a_0.b_0=y\in S.$ Hence, $S$ is closed.

Answer 1

Hence, $b_{n_k}$ converges to $b_0 \in B$(say).

Can you justify this?

Note also that you never used the hypothesis that $B$ is closed; is this hypothesis truly unnecessary?

Answer 2

It might help your proof to clarify that, since $y_n\to y$, then $y_{n_k}\to y$ for any subsequence $n_k$, in which case the sequence $b_n$ has a convergent subsequence $b_{n_k}=y_{n_k}/a_{n_k}\to y/a_0$ (using the subsequence for which $a_{n_k}\to a_0\in A$). Now, since $B$ is closed, we have $y/a_0\in B$, so we can call it $b_0$.