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Show that the norm-closed unit ball of $c_0$ is not weakly compact; recall that $c_0^*=\ell_1$.

Tom Cooney
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Ali Qurbani
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  • Did you mean $C_0$ instead of $c_0$? – Zev Chonoles May 19 '12 at 08:11
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    @Zev: What is $C_0$? I'm pretty sure $c_0$ here means the space of sequences converging to $0$, and that is standard notation. On the other hand, "$C_0$" is often followed by a topological space to denote the space of continuous complex-valued functions on that space that vanish at infinity. E.g. $C_0(\mathbb N)=c_0$ (when $\mathbb N$ is given the discrete topology). – Jonas Meyer May 19 '12 at 21:15

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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 $z=(1,1,\ldots)$.

David Mitra
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  • In general, $B(X)$ is weakly compact if and only if $X$ is reflexive. Additionally, Eberlein-Smulian Theorem tells us that in this case $B(X)$ is weakly sequentially compact. – Theo May 18 '12 at 23:35
  • @DavidMitra I am sorry I am new to analysis. Can you give a detailed answer. Thank you! – Answer Lee Mar 18 '18 at 23:35
  • @AnswerLee write $z=(z_n)$, take $e_m=(0,...,0,1,0,...)$ and note for every subsequence $x_{n_k}$ we must have $$z_n = \langle z,e_m\rangle = \lim_n\langle x_{n_k},e_m\rangle=1$$ Then, $z_n\not\in c_0$ – Veridian Dynamics Mar 18 '18 at 23:44
  • @VeridianDynamics I am sorry I don't get that why $$z_n = \langle z,e_m\rangle = \lim_n\langle x_{n_k},e_m\rangle=1?$$ Can you be more specific? Thank you!! – Answer Lee Mar 19 '18 at 00:22
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    for sequences $(x_n)\in c_0$ and $(y_n)\in(c_0)^*=l^1$, the value of the functional $(y_n)$ on the point $(x_n)$ is given by: $$\langle (x_n),(y_n)\rangle = \sum_{n=1}^\infty x_ny_n$$ So, for every $n_k$ larger than $m$, we have $\langle x_{n_k},e_m\rangle = 1$. This gives $\lim_n\langle x_{n_k},e_m\rangle=1$. $\langle z,e_m\rangle = \lim_n\langle x_{n_k},e_m\rangle$ follows from the continuity of $z$ – Veridian Dynamics Mar 19 '18 at 00:39
  • @VeridianDynamics Thanks for your answer. Can you explain why $z_n = \langle z,e_m\rangle$? – Answer Lee Mar 20 '18 at 18:52
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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.

Per Mattson
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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.

Per Mattson
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