I am finding serious difficulties in understanding some things about relative countable compactness and the use of sequences in proving it by my functional analysis text, Kolmogorov-Fomin's.

For example, here in corollary 2, it says that a subset of the space $E^{\ast}$ conjugate to a separable Banach space $E$ is bounded if it is relatively countably compact (as in definition 5 here) in the weak$^{\ast}$ topology as a consequence of theorem 2' here. I would like to understand why, if $M$ is not bounded, it cannot be relatively countably compact, but from theorem 2' I only get that, if $\{f_n\}_n$ is not bounded, then it is not weakly$^{\ast}$ convergent: how can that prove that any infinite subset of $M$ has an accumulation point?

My book, without giving a proof and treating it as trivial, might appear to treat countable compactness in the weak$^{\ast}$ topology as equivalent to the fact that any sequence has a a weak$^{\ast}$-convergent subsequence, but, if it is true, I am too stupid to see it as trivial, though I am not sure that it is true...

If $M$ were a subset of a metric space I would know that $x$ would be an accumulation point of $M$ if and only if it were the limit of an eventually non-constant sequence of points belonging to $M$, but $E^{\ast}$ with the weak$^{\ast}$ topology is not a metric space (though a sphere centred in 0 is metrisable, but, if we do not a priori know that a subset $M$ is bounded...).

I also think, thanks to what, and to who has written what, I read here, that if $E$ is a separable Banach space then every relatively weak$^{\ast}$-compact subset of $E^{\ast}$ is relatively sequentially weak$^{\ast}$-compact (definition as here), but here we only have countable weak$^{\ast}$-compactness... Does relative countable weak$^{\ast}$-compactness implies relative sequential weak$^{\ast}$-compactness? If it does, I see that if $M$ is not bounded, and infinite, we can chose from it a sequence $\{f_n\}$ (even such that $\forall i\ne j\quad f_i\ne f_j$, if we desire so) such that $\|f_n\|\to+\infty$, which, by theorem 2', will not have any convergent subsequence. Then, if relative countable weak$^{\ast}$-compactness implied relative sequential weak$^{\ast}$-compactness, $M$ would not be countabye weak$^{\ast}$-compact, so that the "only if part" of corollary 2, which is what causes some problems to me, would be proven. But I am far from being conviced that such implication is true...

I uncountably thank you for any help! ;-)


I agree that the book should have explained Corollary 2 a bit better, just as you explained.

Nevertheless, we can prove Corollary 2 by contradiction (as far as I noticed, this book does not use Banach-Steinhauss/Uniform Boundedness Principle, which I will use): Suppose $M$ is an unbounded rcc (relatively countably compact). Let $(f_n)_{n=1}^\infty\subseteq M$ with $\Vert f_n\Vert\to\infty$. Since $(f_n)$ is not norm-bounded in $E^*$, the Uniform Boundedness Principle implies that it is not pointwise-bounded, so there exists $x\in E$ with $\sup_{n\in\mathbb{Z}_{>0}}|f_n(x)|=\infty$. Taking a subsequence if necessary, suppose $|f_n(x)|\geq n$.

But $M$ is rcc, so $(f_n)$ has a limit point $F$. The set $U=\left\{g\in E^*:|g(x)-F(x)|<1\right\}$ is a weak*-neighbourhood of $F$, so $U$ contains infinitely many $f_n$'s, so, for infinitely many $n\in\mathbb{Z}_{>0}$, $|f_n(x)-F(x)|<1$, hence $n-1\leq |f_n(x)|-1\leq|f_n(x)-F(x)|+|F(x)|-1<|F(x)|$, a contradiction.

Alternatively, I believe you can proceed with a proof similar to that of Theorem 2.

Your comment on rcc implying sequential weak*-compactness is true. Moreover, we can prove that the following are equivalent, given $M\subseteq E^*$, with $E$ separable:

  1. $M$ is a closed rcc.
  2. $M$ is sequentially weak*-compact.
  3. $M$ is bounded and weak*-closed.
  4. $M$ is weak*-compact.

(1.$\Rightarrow$2.) Suppose $M$ is a closed rcc. Let $(f_n)\subseteq M$ be an arbitrary sequence. By Corollary 2, $(f_n)$ is bounded, so Theorem 4 implies that $(f_n)$ has a weak*-convergent subsequence. Therefore $M$ is sequentially weak*-compact.

(2.$\Rightarrow$3.) Suppose $M$ is sequentially weak*-compact. First let's show that $M$ is bounded: Suppose not. Then, by Uniform Boundedness, $M$ is not pointwise bounded, so there is $y\in E$ and $(f_n)\subseteq M$ such that $\Vert f_n(y)\Vert \to\infty$. By sequentially weak*-compactness, we can assume, going to a subsequence if necessary, that $f_n$ converges weakly* to some $F\in E^*$. But then $|F(y)|=\lim |f_n(y)|=\infty$, a contradiction. Hence $M$ is bounded.

Now we show that $M$ is closed. Let $(x_i)\subseteq E$ be a dense sequence. Suppose $F\in\overline{M}$ (the closure in the weak*-topology). For every $n$, choose $f_n\in M$ such that $|f_n(x_i)-F(x_i)|<1/n$ for every $i=1,\ldots,n$. In particular $f_n(x_i)\to F(x_i)$ as $n\to\infty$ for every $i$. Going to a subsequence, assume $f_n$ converges weakly* to some $G\in M$. Then $G(x_i)=\lim f_n(x_i)=F(x_i)$ for every $i$. Since $(x_i)$ is dense, $F=G\in M$, so $M$ is closed.

(3.$\Rightarrow$4.) If $M$ is bounded and closed, then it is a closed subset of some closed ball in $E^*$, which is weak*-compact by Banach-Alaoglu, so $M$ is weak*-compact itself.

(4.$\Rightarrow$1.) If $M$ is weak*-compact (in particular it is closed) and $S$ is an infinite subset of $M$, let $(f_n)\subseteq S$ be an infinite subsequence. By compactness, $(f_n)$ has a subnet which converges to some $F\in E^*$, and $F$ is clearly a limit point of $S$, so $M$ is rcc.

  • $\begingroup$ It's a pity that the maximum score is +1. You would deserve $n$ where $n\to\infty$ :-) I admit that I don't perceive the proof "immediate" as the book says. Banach-Steinhaus theorem hasn't even been proven nor stated in the book at this point. You saved me from being blocked in a situation where my studies couldn't go further. $\aleph_1$ graças!!! $\endgroup$ – Self-teaching worker Sep 15 '14 at 16:04

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