Proving that a set is closed Let $A$ be a compact subset of $\mathbb{R}-\{0\}$ and $B$ be a closed subset of $\mathbb{R}^n$.
Prove that the set $\{ a \times b\colon a \in A, b \in B\}$ is closed in $\mathbb{R}^n$.
I am planing to show that it contains all its limit points.
But how do I show that?
 A: you have to notice that every compact set in an hausdorff space is closed. Since $\mathbb{R}$ is a hausdorff space, it follows, that $A$ is compact and so the product $A \times B \subset \mathbb{R}^{n+1}$ is also compact
A: If $A$ is compact in an Hausdorff space then  $A$ is closed. Now $A$ and $B$ are closed sets. Then $A\times B\subset\mathbb{R}^{n+1}$ is closed. 
A: For brevity write $AB = \{ab : a \in A, b \in B\}$.
Suppose that $x$ is a limit point of $AB$. We want to show that $x \in AB$. Since $x$ is a limit point, there exists a sequence $(a_n b_n)$ where $a_n \in A$ and $b_n \in B$, such that $a_n b_n \to x$.
As $A$ is compact, $a_n$ has a subsequence, say $(a_{n_k})$, that converges to some point $a \in A$. (Note that this means that $a \neq 0$.)
Observe that $(a_{n_k} b_{n_k})$ is a subsequence of $(a_n b_n)$, hence also converges to $x$.
Since $a_{n_k}$ is always nonzero, we can form the quotient 
$$b_{n_k} = \frac{a_{n_k}b_{n_k}}{a_{n_k}} \to \frac{x}{a}$$
In other words, $b_{n_k}$ converges to the limit $b = x/a$, which is in $B$ because $B$ is closed.
Thus we have established that $x = a(x/a) = ab \in AB$ as desired.
