# How the norm-closed unit ball of $c_0$ is not weakly compact?

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!!

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.

• 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? Mar 19, 2018 at 0:28
• @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. Mar 19, 2018 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! Mar 19, 2018 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. Mar 19, 2018 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. Mar 19, 2018 at 16:43