My question is how to prove that the following cover does not have a finite sub cover :
$$A_{n}=\{x \in \mathbb{R}^{n}: \frac{1}{2n}<|x|<\frac{3}{2n} \}$$
for the pointed ball $B_1^*(x) = \{ x: 0 < |x| \le 1 \}$
But the thing is how can I prove it because I don't know is there is a standard method to do it or how do I have to proceed, I think that this has to be easy, but I need your help :) thanks in advance
Suppose that $U = \{ A_i \mid i \ J \}$ is a subcover, where $J$ is a set of integers. Every subcover can be written this way: you just include in $J$ the indices of all the "A"-sets that are included.
If $J$ were finite, then it has a maximum (every finite set of integers has a maximum element); class this maximum $N$. Then the point $(\frac{1}{4N}, 0, \ldots, 0)$, is not covered by $U$. Why not? Suppose that it were. Then for some $i \in J$, we'd have $$ \frac{1}{2i} < \frac{1}{4N} < \frac{3}{2i} $$ Ignoring the second inequality, this says that
$$ \frac{1}{2i} < \frac{1}{4N} \\ 2i > 4N \\ i > 2N $$ But since $N$ is the maximum of a set of positive integers, it's positive. Hence $2N > N$, so we conclude $i > N$, which is a contradiction, since $N$ was defined to be the largest element of $J$.
(By the way, the letter $n$ is used for two different things in this problem. (1) the dimension of the space $\mathbb R^n$, and (2) an index for elements of an open cover. I suggest replacing the first use by $k$.)