# Let $A , B\subseteq\mathbb{R}$. If $A$ is closed and $B$ is compact, is $A\cdot B$ closed?

Let $A , B\subseteq\mathbb{R}$. If $A$ is closed and $B$ is compact, is $A\cdot B=\{a. b: a\in A,b\in B\}$ closed?

If $p$ is an adherent point in $A\cdot B$, exists a sequence $p_{n}=\alpha_{n} \beta_{n}$, $\alpha_{n} \in A$ and $\beta_{n}\in B$, such that $p_{n}\rightarrow p$. If $B$ is compact, $\beta_{n}\rightarrow b \in B$.

#### Solutions Collecting From Web of "Let $A , B\subseteq\mathbb{R}$. If $A$ is closed and $B$ is compact, is $A\cdot B$ closed?"

Let $A=\mathbb{N}$ and $B=\{\frac{1}{n}:n \in \mathbb{N} \}\cup\{0 \}$ , then A is closed and B is compact but $A.B$ is no closed because $\sqrt{2}$ is an adherent point of $A.B$ , and $\sqrt{2} \notin A.B$ .

$C=A\times B$, catch hold a sequence $c_n=a_n\times b_n$ such that $c_n\to c$

we need to show $c=a\times b$ for some $a\in A, b\in B$

since $b_n\in B$ and $B$ is closed, bounded so it has a convergent subsequence say $b_{n_k}\to b$

now $a_n={c_n\over b_n}\in A, a_n\to {c\over b}$ but $A$ is closed so ${c\over b}\in A$, call it $a$

so $c=a\times b$ Done!

In harmonic analysis we have if $G$ is topological group and $A$ and $B$ be subset of $G$ , $A$ closed and $B$ is compact then $A.B$ is closed.(proof in principle of harmonic analysis written by anton Deitmar , Lemma 1.1.4). and compact condition is needed i.e. if A and B are closed then AB is not closed.

If we add the condition $0 \not \in B$, I think that the answer is yes.