Things that we know:

  1. In any topological space compactness implies sequential compactness
  2. 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!

  • 2
    $\begingroup$ Compactness does not in general imply sequential compactness. $\endgroup$ – David Mitra Jun 3 '14 at 18:22
  • $\begingroup$ There is a proof in the Lang's book, Real and Functional Analysis, books.google.com.br/books/about/… in page 33 that say the opposite $\endgroup$ – O Empalador de Cabras Jun 3 '14 at 18:30
  • $\begingroup$ Compactness implies sequential compactness in metrizable spaces. But the weak* topology on $X^*$ is not metrizable in general (in particular, if $X$ is Banach and infinite dimensional). $\endgroup$ – David Mitra Jun 3 '14 at 18:32
  • $\begingroup$ Please ignore my, now deleted, previous comment. $\endgroup$ – David Mitra Jun 3 '14 at 18:35
  • 1
    $\begingroup$ That a sequence has an accumulation point is strictly weaker than that it has a convergent subsequence (in general; for metric spaces, more generally first countable spaces, the two are equivalent). $\endgroup$ – Daniel Fischer Jun 3 '14 at 18:46

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.

  • $\begingroup$ Another condition perhaps worth mentioning is that the unit ball of $X^*$ is weak* sequentially compact if $X^*$ does not contain an isomorphic copy of $\ell_1$. This follows from Rosenthal's $\ell_1$-Theorem (A Banach space $X$ fails to contain an isomorphic copy of $\ell_1$ if and only if every bounded sequence in $X$ has a weakly Cauchy subsequence). $\endgroup$ – David Mitra Jun 3 '14 at 22:40

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