Weak* compactness of the unit ball Things that we know:


*

*In any topological space compactness implies sequential compactness

*If E is any topological space the then the closed unit ball 
$$
B_E=\{f\in E^*; \|f\|\leq 1\}
$$
is compact in the weak* topology.


Now an example of Brezis Book: Let $E=l^{\infty}$ and its dual $E^*\supset l^1$.  Now consider the sequence $(f_n)\subset l^1\subset E^*$ where 
$$
f_n=(0, \ldots,0,1,0\ldots)
$$

Claim.:  $(f_n)$ has no convergent subsequence in the weak* topology

Arguing by contradiction, suppose there is a convergent subsequence $f_{n_k}$  converging to $ f $. So we must have for any $x\in l^{\infty}$ that 
$$
\left<f_{n_k},x\right>\to \left<f,x\right>
$$
On the other hand choose $x_0$ in the following way
$$
x=(0,0,\ldots,\underbrace{1}_{n_1}, 0,0\ldots,0,\underbrace{-1}_{n_2},0,0,\ldots,\underbrace{1}_{n_3},0,\ldots,0,\underbrace{-1}_{n_4}, \ldots)
$$
then 
$$
\left<f_{n_k},x\right>=(-1)^k
$$
which does not converge, contradiction! So  $(f_n)$ has no convergent subsequence in the weak* topology
MY QUESTION: How this example does not contradict the results $1$ and $2$, 
I mean, $\|f_n\|=1$ for all $n$, and  $(f_n)$ has no convergent subsequence in the weak* topology. 
On the other hand $B_E=\{f\in E^*; \|f\|\leq 1\}$
is compact in the weak* topology, which means in particular, sequentially compact.
ADDENDUM: compactness implies sequential compactness
Let X be a compact set and $(x_n)$ a infinite sequence(infinite distinct terms), suppose the opposite i.e that $(x_n)$ does not have a accumulation point. Then for each $x\in X$ there is a neigh. $U_x$ of $x$ but containing only a finite number of elements of $\{x_n\}$. The family $\{U_x\}$ cover $X$, by passing to a finite subcover $\{U_1,\ldots,U_n\}$ we conclude that the set of the terms of $(x_n)$ must be finite. Contradiction!
 A: Compactness does not imply sequential compactness.
Compactness implies that every sequence has an accumulation point, which is equivalent to countable compactness [every countable open cover has a finite subcover]. But in general, a sequence having accumulation points does not imply that the sequence has a convergent subsequence. One needs additional hypotheses, e.g. first countability of the space to have that implication.
One example of a space that is compact but not sequentially compact is, as shown by the example, the closed unit ball of $(\ell^\infty)^\ast$ in the weak$^\ast$ topology.
A perhaps easier to visualize example is a product of sufficiently many copies of $\{0,1\}$. (Any example must be somewhat difficult to visualize, since the easy-to-visualize spaces have a strong tendency to be first-countable.)
Let $\mathscr{P}(\mathbb{N})$ denote the power set of $\mathbb{N}$, and $X = \{0,1\}^{\mathscr{P}(\mathbb{N})}$ (that is up to the naming of the indices $\{0,1\}^{\mathbb{R}}$, but taking $\mathscr{P}(\mathbb{N})$ makes it easier to define a sequence without convergent subsequences). Define the sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ by
$$p_M(x_n) = \begin{cases} 0 &, n \notin M\\
0 &, n\in M \text{ and } \operatorname{card} \{m\in M : m  < n\} \text{ even}\\
1 &, n\in M \text{ and } \operatorname{card} \{ m\in M : m < n\} \text{ odd},\end{cases}$$
where $p_M \colon X \to \{0,1\}$ is the coordinate projection. Then $(x_n)_{n\in\mathbb{N}}$ has no convergent subsequences. For if $(x_{n_k})_{k\in\mathbb{N}}$ is a subsequence, consider the set $M = \{ n_k : k\in\mathbb{N}\}$. Then $p_M(x_{n_k})$ is $0$ for even $k$ and $1$ for odd $k$ (if you follow the convention $0\notin \mathbb{N}$, switch even and odd), so $(x_{n_k})$ is not convergent.

If $E$ is a normed space, then the closed unit ball of $E^\ast$ is compact in the weak$^\ast$ topology by the Banach-Alaoglu theorem, and under some conditions on $E$ it is also sequentially compact.


*

*If $E$ is separable, then the subspace topology induced on the closed unit ball of $E^\ast$ by the weak$^\ast$ topology is metrisable (Note: The weak$^\ast$ topology on $E^\ast$ is then generally not metrisable itself), hence the closed unit ball of $E^\ast$ is then weak$^\ast$-sequentially compact.

*If $E$ is reflexive, the closed unit ball of $E^\ast$ is weak$^\ast$-sequentially compact.


$\ell^\infty$ is neither separable nor reflexive.
