Show that $A$ and $A^C$ are both dense in $(\ell^2,\lVert \cdot \rVert_2)$, where $A=\{x\in\ell_2:\sum_{k=1}^\infty x_k\neq0\}$.

Question

The title says it all. Showing $A$ is dense in $\ell_2$ seems easy; for any $x\notin A$, for each $n\in\mathbb N$ let $x^n$ in $\ell_2$ where $x^n$ is identical to $x$ except that $x^n_1=x_1 + \frac1n$. Then $x^n\to x$ (right?).

The harder part to me is showing $A^C$ dense in $\ell_2$. I don't really know where to start here. I'm very surprised the statement is even true. I can't wrap my head around how a sequence $\{x^n\}\subseteq \ell_2$ such that $\sum_{k=1}^\infty x^n_k=0\ \forall n$ could possibly converge to, e.g., $x\in\ell_2$ where $\sum_{k=1}^\infty x_k=50$. Am I completely misunderstanding the question? Any pointers?

Answer 1

Part 1. $A$ is dense in $\ell^2$

Your approach of this part is right. What is noteworthy is that, there is at most one of element in $\{x^n\}$ that does not belong to $A$.

Part 2. $A^c$ is dense in $\ell^2$

We need to show that, $\forall x\in\ell^2$, $\exists \{x^n\}\subset A^c$, s.t. $x^n\xrightarrow{\ell^2}{x}$ ($n\to\infty$).
Now for $x=(x_1,x_2,\cdots)\in\ell^2$, let $S_n:=\sum_{k=1}^nx_k$ be the partial sum of $x$. Define $$ x^n:=\Big(x_1,\cdots,x_n,\underbrace{-\frac{S_n}{n^2},\cdots,-\frac{S_n}{n^2}}_{n^2},0,0,\cdots\Big) $$ where the number of term $-\frac{S_n}{n^2}$ is $n^2$, obviously $x^n\in A^c$ for all $n$.
Also let $y^n:=(x_1,\cdots,x_n,0,0,\cdots)$ be the truncation of $x$. We compute \begin{align} \|x-x^n\|_{\ell^2}\le\|x-y^n\|_{\ell^2}+\|y^n-x^n\|_{\ell^2}; \end{align} when $n\to\infty$, $x\in\ell^2$ implies $$ \|x-y^n\|_{\ell^2}=\sum_{k=n+1}^\infty |x_k|^2\to0, $$ and \begin{align} \|y^n-x^n\|_{\ell^2} &= n^2\cdot\bigg(\frac{S_n}{n^2}\bigg)^2=\frac{S_n^2}{n^2}=\frac{1}{n^2}\bigg(\sum_{k=1}^nx_k\bigg)^2 \\ &\le \frac{1}{n^2}\cdot n\sum_{k=1}^nx_k^2 \\ &= \frac{1}{n}\sum_{k=1}^nx_k^2 \\ &\to 0, \end{align} of which the antepenult step use the fact that $(\sum_{k=1}^na_k)^2\le n\sum_{k=1}^na_k^2$.
Hence $\|x-x^n\|_{\ell^2}\to0$ ($n\to\infty$). We are done.

Answer 2

Hint: Take $y\in l_2$ and $N$ such that $\|y-(y_1,y_2,\ldots,y_N,0,0\ldots)\|_{l_2}\leq\epsilon,$ and set $K=\sum_{i=1}^Ny_i.$ Then estimate $$\|(y_1,y_2,\ldots,y_N,0,0\ldots)-(y_1,y_2,\ldots,y_N,-K/M,-K/M,\ldots,-K/M,0,0,\ldots)\|_{l_2}$$ for a large $M$, where the second vector has $M$ coordinates with values $-K/M$, and so is in $A^c$.

Answer 3

The reason it's possible for $A^c$ to be dense is that $Sx=\sum x_n$ is not a bounded linear functional on $\ell^2$. So you can have sequences with zero sum converging in $\ell^2$ to a sequence with non-zero sum; no contradiction because $S$ is not continuous.

In fact, since $\sum x_n$ need not converge for $x\in\ell^2$ it seems to me that the problem is very poorly stated. The definition of $A$ should be

"the set of $x\in\ell^2$ such that $\sum x_n$ does not converge to $0$",

or

"the set of $x\in\ell^2$ such that $\sum x_n$ converges to something non-zero".

The two are different. (If the original problem was stated as one of those two alternatives and you translated it to what you posted you should have posted the original...)

Exactly how you solve the problem would depend on which definition of $A$ is intended.

Of course both versions are true, and easy.

Lemma Suppose $\epsilon>0$. There exists $x\in\ell^2$ with $||x||_2<\epsilon$ such that $\sum x_n$ diverges.

Proof $\sum_{n=N}^\infty 1/n$ diverges, while $\sum_{n=N}^\infty1/n^2<c/N$.

Lemma Suppose $\epsilon>0$ and $\alpha\in\Bbb C$. There exists $x\in\ell^2$ such that $||x||_2<\epsilon$ and $\sum x_n=\alpha$.

Proof $\sum_{j=1}^N\alpha/N=\alpha$ and $\sum_{j=1}^N|\alpha/N|^2=|\alpha|^2/N$.

Now the two versions of the problem give two different sets $A$; with their complements there are four sets we need to show are dense:

$A$: converges to $0$

$B$: does not converge to $0$

$C$: converges to something non-zero

$D$: does not converge to something non-zero.

(In fact of course $A\subset D$ and $C\subset B$ so it's enough to show $A$ and $C$ are dense.)

Prop Each of the four sets above is dense in $\ell^2$.

Proof Suppose $x\in\ell^2$. Let $s_N=(x_1,\dots,x_N,0,0,\dots)$. So $s_N\to x$, and so if $||t_N||_2<1/N$ then $s_N+t_N\to x$. The lemmas allow you to choose $t_N$ so $s_N+t_N$ is in whichever of $A$, $B$, $C$ or $D$ you want.