I have to prove that a closed unit ball in $C[0,1]$ is not weak-compact. The hint is that I should consider sets:
$$V_t=\{f\in C[0,1]:|f(t)|>1/3\}$$
and
$$U_t=\{f\in C[0,1]:|f(t)|<2/3\}$$
Now I should show that $$\{V_t:t\in \mathbb{Q}\cap[0,1]\} \cup \{U_t:t\in (\mathbb{R} \setminus \mathbb{Q})\cap[0,1]\}$$
is an open cover of a closed unit ball in weak topology such that we can't choose a finite subcover.
Weak topologies are sth new to me. Can anyone help me? I did only manage to show, that $\phi_t:C[0,1]\ni f \rightarrow f(t)\in K$ are bounded linear functionals.
Best Answer
As you said, $\phi_t$ is a bounded linear funcional, so as
$$V_t = \phi_t^{-1} \big( \{ |x| > \frac 13 \}\big), $$
then $V_t$ is open in the weak topology. Similarly $U_t$ is also open. It is quite obvious that
$$\{V_t:t\in \mathbb{Q}\cap[0,1]\} \cup \{U_t:t\in (\mathbb{R} \setminus \mathbb{Q})\cap[0,1]\}$$
is an open cover of the unit ball in $C[0,1]$ (Actually, it is an open cover of the whole $C[0,1]$).
Now we show that: for any finite set $t_1, \cdots t_m \in \mathbb Q\cap [0,1]$, $s_1, \cdots s_n \in \mathbb R\setminus \mathbb Q \cap [0,1]$, there is a function $f \in C[0,1]$ such that $||f||\leq 1$ and
$$f\notin \bigcup_{j} V_{t_j} \cup \bigcup_{k} U_{s_k}$$
Well, it just say that I can find a continuous function $f$ such that $|f(t_j)|\leq 1/3$ (so not in $V_{t_j}$) and $|f(s_k)| \geq 2/3$ (so not in $U_{s_k}$) and $||f|| \leq 1$. Such a function can be found by setting
$$f(t_j) =1/3, \ \ \ \ f(s_k) = 2/3$$
and joining line along adjacent $t_j$, $s_k$. This shows that we cannot pick a finite subcover to cover the closed unit ball in $C[0,1]$, thus the unit ball is not weakly compact.