Show that $A$ is compact implies $A \cdot [0, 1]$ is compact Let $A \subset \mathbb{R}^p$ a compact set. Show that $B = [0, 1] \cdot A = 
\left \{ \alpha a \mid a \in A,\, 0 \le \alpha \le 1 \right\}$ is compact set.
Proving that $B$ is bounded is fairly easy: since $A$ is bounded, there exists a $M > 0$ with $\|a\| \le M,\ \forall a \in A$, so it follows that
$\|\alpha a\| \le 1\cdot M,\ \forall \alpha a \in B $.
I can't figure out how to prove that $B$ is a closed set.
This is what I have done:
Using the sequences definition we should prove that for all sets
$ (b_n)_{n\in\mathbb{N}} \subset B$ with $\lim\limits_{n\to\infty} b_n = b$ we have $b \in B$.
We can write $b_n = \alpha_n a_n$, where $a_n \in A,\ \alpha_n \in [0, 1]$.
The problem is that we don't know if $(a_n)_{n\in\mathbb{N}}$ and $(\alpha_n)_{n\in\mathbb{N}}$ are convergent.
If that were the case we could simply say that $b=\alpha a$, where $\alpha$ and $a$ are the limits of the sequences $(\alpha_n)$ and $(a_n)$ respectively, and since $a \in A$ and $\alpha \in [0, 1]$ we have that $b \in B$.
But I found a counterexample of a convergent series that is the product of two divergent sequences:
Let $A = [0, 1]$, $a_n = \begin{cases} 0,& n\text{ even}; \\ 1, & n\text{ odd,}\end{cases} $ $\quad \alpha_n = \begin{cases} 1,& n\text{ even}; \\ 0, & n\text{ odd.}\end{cases} $
Here we have that $b_n = 0,\ \forall n \in \mathbb{N}$,
$a_n $ and $\alpha_n$ are divergent and the limit of $(b_n)_{n\in\mathbb{N}}$, namely $0$, is a member of $B$.
Is it true that all convergent sequences from $B$ can be written as the product of convergent sequences from $[0, 1]$ and $A$?
 A: An alternative approach. Consider $f:[0,1]\times A\rightarrow\mathbb{R}^p$ defined by $f(a,x)=ax$, $f$ is continuous and its image is compact since the source is compact (Tichonoff: the product of compact sets is compact).
A: Here is a way to continue your proof. Basically the answer to your question is that you can write a subsequence of $(b_n)$ as a product of convergent sequences.
First, since $A$ is compact, there is a convergent subsequence of $(a_n)$. Since the notation gets messy I will write this subsequence as $(a_n)_{n\in I}$ where $I$ is an infinite subset of $\mathbb{N}$. Let $a$ be the limit of this subsequence.
Next consider $(\alpha_n)_{n\in I}$. Since $[0,1]$ is compact it has a convergent subsequence $(\alpha_n)_{n\in J}$ for some infinite $J\subseteq I$. Let $\alpha$ be the limit of this subsequence. Note that $(a_n)_{n\in J}$ still converges to $a$.
Now $b=\lim_J b_n=\lim_J a_n\alpha_n=\lim_J a_n \cdot \lim_J \alpha_n= a\cdot \alpha\in A\cdot [0,1]$.
A: Check that if $*U1,U2,..,Uk*$ cover $A$, then $B*U1,B*U2,...,B*Uk$ cover $B*A$.
