Consider $\ell^\infty $ the vector space of real bounded sequences endowed with the sup norm, that is
$||x|| = \sup_n |x_n|$ where $x = (x_n)_{n \in \Bbb N}$.
Prove that $B'(0,1) = \{x \in l^\infty : ||x|| \le 1\} $ is not compact.
Now, we are given a hint that we can use the equivalence of sequential compactness and compactness without proof.
However, I don't understand how sequences of sequences work? Do I need to find a set of sequences and order them such that they do not converge to the same sequence?
Does the sequence $(y_n)$ where $y_n$ is the sequence such that $y_n = 0$ at all but the nth term where $y_n =1$ satisfy the requirement that it does not have a convergent subsequence?
I think I have probably just confused myself with this sequence of sequences lark. Sorry and thanks.
Best Answer
You have to find a sequence in $B'(0,1)$ which does not contain a convergent subsequence. Because the existenc of such a sequence implies (right from the definition) that $B'(0,1)$ isn't sequentially compact.
Hint Consider the sequence $(y^n)_n$ defined by $$y^n_k := \begin{cases} 1 & n=k \\0 & n \not= k \end{cases} \qquad (k \in \mathbb{N})$$ Then $y^n \in B'(0,1)$ for all $n \in \mathbb{N}$ and $\|y^n-y^m\|_{\infty}=1$ for $m \not= n$. This means that $(y^n)_n$ doesn't contain a Cauchy-subsequence, hence in particular no convergent subsequence.