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I saw an answer here: The norm-closed unit ball of $c_0$ is not weakly compact.

But I am new to analysis. I couldn't understand this answer. All I know is someone is using net accumulation to get compactness. But I don't know why the net doesn't have a weak accumulation point.

Can anyone give me a detailed answer on this question?

Thank you so much!!

Answer Lee
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1 Answers1

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Say $x_n$ is the sequence consisting of $n$ ones followed by all zeroes, as in that post. Let $E_n=\{x_n,x_{n+1},\dots\}$.

Then $E_n$ is weakly closed (details below). If the unit ball of $c_0$ were weakly compact then $E_n$ would be weakly compact. But that's impossible, since the intersection of any finite number of $E_n$ is non-empty but the intersection of all the $E_n$ is empty.

Details regarding why $E_n$ is weakly closed:

Given $x\in c_0$, a finite set $F\subset\Bbb N$, and $\epsilon>0$ let $V(x,F,\epsilon)$ be the set of all $x'\in c_0$ such that $|x'(j)-x(j)|<\epsilon$ for every $j\in F$. Then $V(x,F,\epsilon)$ is weakly open, because for each $j$ the map $x\mapsto x(j)$ is weakly continuous.

Now note that $x\in c_0$ is an element of $E_n$ if and only if (i) $x(j)$ is $0$ or $1$ for every $j$, (ii) $x(j)=1$ for all $j\le n$ and (iii) if $x(j)=0$ and $k>j$ then $x(k)=0$. Hence if $x\notin E_n$ there exist $F$ and $\epsilon$ such that $V(x,F,\epsilon)\cap E_n=\emptyset$. So the complement of $E_n$ is weakly open.

David C. Ullrich
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  • I am sorry I don't get the part the intersection of any finite number of $E_n$ is non-empty but the intersection of all the $E_n$ is empty. And I know $E_n$ would be weakly compact. How to get $E_n$ is not weakly compact? – Answer Lee Mar 19 '18 at 00:28
  • @AnswerLee Have you really thought about this? $\bigcap_{j=1}^N E_n=E_N\ne\emptyset$. And $E_n={x_m:m\ge n}$, so $\bigcap E_n=\emptyset$. Hence the $E_n$ are not weakly compact. That's a basic property of compactness: If the $E_n$ are compact and the intersection of any finite number of them is nonempty then the intersection of all of them is nonempty - if that's not familiar you need to learn a little elementary topology. – David C. Ullrich Mar 19 '18 at 14:53
  • I understand it now. But I saw you edit your answer a little bit. I am wondering is $E_n$ normed closed in $c_0$? And using $E_n$ is normed closed in $c_0$ to get $E_n$ is weakly closed. Thank you! – Answer Lee Mar 19 '18 at 16:38
  • @AnswerLee The reason for the edit was I realized that "norm closed implies weakly closed" was nonsense. It's clear from the definitions that in fact weakly closed implies norm closed, because weakly open implies norm open. – David C. Ullrich Mar 19 '18 at 16:41
  • @ DavidC.Ullrich No I think it is true because this theorem: Let $X$ be a normed space. Then a convex subset of $X$ is closed if and only if it is weakly closed. – Answer Lee Mar 19 '18 at 16:43
  • @AnswerLee ??? $E_n$ is obviously not convex. – David C. Ullrich Mar 19 '18 at 16:47
  • We can also show that the set $E$ of all binary sequences in $c_0$ is a closed discrete subspace of the norm-closed unit ball in the weak topology. Any space with an infinite closed discrete subspace cannot be a compact space – DanielWainfleet Mar 19 '18 at 16:53
  • @DavidC.Ullrich Yes you are right. Thank you for pointing out. – Answer Lee Mar 19 '18 at 16:54
  • @DavidC.Ullrich I am sorry if I have too many questions. What is your $x(j)$ and why is $x\mapsto x(j)$ weakly continuous? Thank you! – Answer Lee Mar 20 '18 at 00:31
  • @AnswerLee The notation "$x_n$" is already taken, so instead of thinking of $x$ as $(x_1,x_2,\dots)$ I'm taking $x=(x(1),x(2),\dots)$ for the generic element of $c_0$. (That map is weakly continuous by definition, since it's an element of the dual space.) – David C. Ullrich Mar 20 '18 at 01:14
  • @DavidC.Ullrich Can you be more specific why the map is an element of the dual space? Thank you so much! – Answer Lee Mar 20 '18 at 01:30
  • @AnswerLee Are you trying to figure these things out? This is literally just the definition. Define $L:c_0\to\Bbb C$ by $Lx=x(j)$. Saying $L\in c_0^*$ means that $L$ is linear and there exists $c$ such that $|Lx|\le c||x||$ for every $x$. Do you want me to explain why $L$ is linear? Or do you want an explanation of why $|x(j)|\le c||x||=c\sup_k|x(k)|$? – David C. Ullrich Mar 20 '18 at 12:41