The norm-closed unit ball of $c_0$ is not weakly compact Show that the norm-closed unit ball of $c_0$ is not weakly compact; recall that $c_0^*=\ell_1$.
 A: Hint: Let $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$. Suppose $z\in c_0$ is a weak cluster point of $(x_n)$. By considering the action of the standard unit vectors of $\ell_1$ on the $x_n$, obtain a contradiction by showing that we must have $z=(1,1,\ldots)$. 
A: The earlier proofs are very elegant, but rely implicitly on the result that compactness implies either limit point compactness (as in the hint by @DavidMitra), or sequential compactness (as in the proof in the comments by @VeridianDynamics). In general, these three notions are not equivalent! They are equivalent for metric spaces, but the weak topology on an infinite-dimensional space is not metrizable. It turns out that these notions are also equivalent in the weak topology on a Banach space, which is the case at hand, due to a theorem by Eberlein and Smulian, but that result is highly non-trivial. That said, we don't need equivalence; we only need the implication that compactness implies limit point compactness, and that can be established by simpler means, but does require an argument. I give a full proof based on the hint of @DavidMitra in another answer. I'd be interested if someone can simplify it.
Here's another proof, which is based directly on the definition that $E$ is compact if every open cover has a finite subcover. For another proof along these lines, see Ullrich.
Let $E = \{e_1,e_2,\ldots\}\subset B$, where $B$ is the unit ball, and where $e_m(n)= 1$ for $n=m$ and zero otherwise. It is easy to see that $E$ is weakly closed. If the unit ball were weakly compact, then $E$ would also be weakly compact. We show that it is not.
Let $V_m = \{x: |x(m) - 1| < \epsilon\}$, with $\epsilon>0$ small. Then  $V_m$ is a weak neighborhood of $e_m$, the $V_m$ are all disjoint, and ${\mathcal V} = \{V_m\}$ is a countable weak open cover of $E$. If any of the $V_m$ are removed, then the corresponding $e_m$'s are not covered. Therefore, ${\mathcal V}$ does not have a finite subcover, $E$ is not weakly compact, and $B$ is not weakly compact.
A: Here is a proof based on @DavidMitra's hint. We give a Lemma showing that if $B$ is compact, then any sequence in $B$ has a weak cluster point. We then show that the sequence $(x_n)$ does not have a weak cluster point, which implies that $B$ is not compact.
We define a cluster point of a
sequence $A$ in a topological space $X$ as a point $x\in X$ such that  any neighborhood $U$ of $x$ contains
an infinite number of points from $A$.
Lemma: If $B$ is compact, then every sequence in $B$ has a cluster
point. (Adapted from Wikipedia entry on Countable Compactness.)
Proof: Let $A = (x_1,x_2,\ldots)$, and assume that $A$ has no
cluster points. No $x$ can occur more than a finite number of times,
or it would itself be an accumulation point. Therefore, $A$ is
infinite. For each $x\in B$, let $U_x$ be a neighborhood of $x$.  For
every finite subset $F$ of $A$, let
$U_F = \bigcup\{U_x: U_x\cap A = F\}$. Since every $U_x$ is a subset of
one of the $U_F$, the $U_F$ cover $X$. Let $\{\tilde U_i\}$ be a
finite subcover.  Then the $\tilde U_i$ cover at most a finite number
of points from $A$ and cannot cover $A$ or, a fortiori, $B\supset A$. Thus, $B$
is not compact.
Proof that $c_0$ is not weakly compact:
Consider the sequence $(x_n)$, where $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$, as above.
Suppose $z$ is a weak cluster point of $(x_n)$. Then an infinite
number of the $x_n$ are in every weak neighborhood of $z$. Let
$$ V_m = \{x:|x(m) - z(m)| < \epsilon\}$$
be a weak neighborhood of $z$. A finite number of $x_n$ have $x_n(m)=0$,
and an infinite number have $x_n(m) = 1$. The only way an infinite number of the $x_n$ can be in $V_m$ is if $z(m)=1$. But $m$ was arbitrary, so
all the entries of $z$ are one, that is, $z = (1,1,\ldots)$, but then
$z\not\in c_0$. Therefore, the sequence $(x_n)$ does not have a cluster point, and by the Lemma, $B$ is not compact.
