Let $X$ bet the set of all sequences $\{a_n\}_{n=1}^\infty$ in
$\mathbb R$ with $\lim_{n \to \infty} a_n = 0$. For any $\{a_n\},
\{b_n\} \subset X$, we define a metric$$d(\{a_n\}, \{b_n\}) = \sup\{|a_n – b_n|, n = 1,2,\dots \}$$
Prove that the unit ball $\{\{a_n\}\in X: d(\{a_n\}, 0) = \sup_{n \in
\mathbb N} |a_n| \le 1\}$ in $X$ is not compact
I think this unit ball is both bounded and closed in metric space, so it is compact. But the question asks me to prove that it is not compact. Please help.
Best Answer
One of the more intuitive definitions/characterizations of compactness in a metric space $(X,d)$ is the following:
A set $K \subseteq X$ is compact if and only if every sequence $\{a_k\} \subseteq K$ has a convergent subsequence.
So in order to show that the unit ball in the space of all sequences is not compact, you have to find a sequence (of sequences) in the unit ball that does not contain a convergent subsequence.