The nature of the subset $\{a\cdot b : a \in A \text{ , } b \in B\}$ where $A$ and $B$ are different subsets of $\Bbb R$ and $\Bbb R^n$ respectively I was trying to solve the following problem : (i) Let $A$ be a compact subset of $\Bbb R - \{0\} $ and $B$ be a closed subset of $\Bbb R^n$ . Prove that the subset $\{a\cdot b : a \in A \text{ , } b \in B\}$ is a closed set in $\Bbb R^n$ .
My attempt :
Firstly, I tried by , for a given $ x \in \Bbb R$ , considering the map $\Bbb R^n \to \Bbb R^n$ by $ b \mapsto xb$ i.e. scalar multiplication, the image set under the map remains closed. So if our $A$ is a finite union of singleton sets, since finite union of closed sets are closed, the image set i.e. $\cup_{x \in A} {\{x\cdot b : b \in B \}}$ , but I' unable to extend this simple argument for a compact set in $\Bbb R$ with arbitrarily many points.
Next, by considering a convergent sequence with entries in $\{a\cdot b : a \in A, b \in B\}$ , (say) $(x_n) \to x$. $(x_n)$ can be rewritten as $x_n = a_n \cdot b_n$ , where $a_n$ and $b_n$ are respectively sequences in $A$ and $B$. Since $A$ is compact, $(a_n)$ has a converging subsequence, say $(a_{n_k}) \to a$ for some $a \in A$. Then $ (b_{n_k}) \to \frac{x}{a}$ and $\frac{x}{a} \in B$, thus $x \in \{a\cdot b : a \in A, b \in B\}$ , so this works! 
After doing this problem myself, I got a linked question but this  is not a duplicate due to the following part : 
I would also like to know what would be the resultant set had :
(ii) $A$ be an open set in $\Bbb R$ and $B$ remained a closed set in $\Bbb R^n$.
(iii) $A$ had just been a closed set in $\Bbb R$ and $B$ remained a closed set in $\Bbb R^n$.
(iv) $A$ be an open set in $\Bbb R$ and $B$ be an open set in $\Bbb R^n$.
(v) $A$ be an open set in $\Bbb R$ and $B$ be a compact set in $\Bbb R^n$.
(vi) $A$ be an compact set in $\Bbb R$ and $B$ also be a compact set in $\Bbb R^n$.
(vii) $A$ be an compact set in $\Bbb R$ and $B$ be an open set in $\Bbb R^n$.
(viii) $A$ had just been a closed set in $\Bbb R$ and $B$ be an open set in $\Bbb R^n$.
(ix) $A$ had just been a closed set in $\Bbb R$ and $B$ a compact set in $\Bbb R^n$.
No idea about (ii), (iii), (viii), (ix).
I guess (iv) should be an open set in $\Bbb R^n$ . My first approach to the probem (i) should work here, since arbitrary union of opens sets is open in $\Bbb R^n$ .
Intuitively seems like (v), (vii) should be  an open set.
I guess my second approach (working with sequences) in problem (i) should work here and (vi) should be a compact set.
Sorry for asking so many questions at a time but I really want to know and learn about them fast.
Thanks in advance for help.
 A: For $A\subset\mathbb{R}$ and $B\subset\mathbb{R}^n$, denote by $C=\{a\cdot b : a\in A, b\in B\}\subset\mathbb{R}^n$.

(ii) $A$ is open and $B$ is closed: $C$ is neither open nor closed
(v) $A$ is open and $B$ is compact: $C$ is neither open nor closed

Set $A=(-1,1)$ and $B=\{(1,y) : 0\leq y\leq1\}\subset\mathbb{R}^2$. Since any point of $B$ is not contained in $C$, $C$ is not closed. It is trivial to check that $C$ is not open. See the figure below.


(iii) $A$ is closed and $B$ is closed: $C$ is neither open nor closed
(i') $A$ is compact (but containing $0$) and $B$ is closed: $C$ is neither open nor closed

Set $A=[-1,1]$ and $B=\{(1,y) : y\in\mathbb{R}\}\subset\mathbb{R}^2$. Then $C=\{(x,y) : x\in[-1,0)\cup(0,1],\ y\in\mathbb{R}\} \cup \{(0,0)\}$ is neither open nor closed.

(iv) $A$ is open and $B$ is open: $C$ is neither open nor closed
(vii) $A$ is compact and $B$ is open: $C$ is neither open nor closed
(viii) $A$ is closed and $B$ is open: $C$ is neither open nor closed

Set $A=(-1,1)$ for (iv), and $A=[-1,1]$ for (vii) and (viii). For all three cases, set $B=\{(x,y) : (x-1)^2+y^2<1\}\subset\mathbb{R}^2$ (as my comment above). It is trivial that $C$ is not closed since $B$ is an open ball. Moreover, $C$ is not open since $(0,0)$ cannot be an interior point of $C$. See the figure below.


(vi) $A$ is compact and $B$ is compact: $C$ is compact

Since the scalar multiplication $\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n$ is continuous (see Continuity of multiplication in a topological (vector) space), the image of the compact set $A\times B$ by the scalar multiplication, that is $C$, is also compact.

(ix) $A$ is closed and $B$ is compact: $C$ is neither open nor closed

Set $A=\mathbb{N}=\{1,2,3,\dotsc\}$, the set of natural numbers, and $B=\{\frac{1}{n}e^{2\pi i/n}=\left(\frac{\cos(2\pi/n)}{n},\frac{\sin(2\pi/n)}{n}\right) : n=2,3,4,\dotsc\}\cup\{(0,0)\}$ which is compact ($\because$ closed and bounded). Then clearly $C$ is not open, and easy to check that $C$ is not closed since the set $\{(x,0) : x>0\}$ is not contained in $C$. See the figure below.

Remark. According to the answers above (from (ii) to (ix)), only the following questions are meaningful:


*

*If both $A\subset\mathbb{R}$ and $B\subset\mathbb{R}^n$ are compact, then so is the set $\{a\cdot b : a\in A, b\in B\}\subset\mathbb{R}^n$.


*If $A\subset\mathbb{R}\setminus\{0\}$ is compact and $B\subset\mathbb{R}^n$ is closed, then the set $\{a\cdot b : a\in A, b\in B\}\subset\mathbb{R}^n$ is closed.

