TITLE: Is $\ell^1 \subset \ell^2$ meagre?
QUESTION [5 upvotes]: Possible Duplicate:
Prove $\ell_1$ is first category in $\ell_2$ 

Consider $\ell^2$ with the topology induced by the usual norm. We can easily prove that $\ell^1 \subset \ell^2$. I am wondering if $\ell^1$ is meagre (i.e. of first category) in $\ell^2$. In other words, I am looking for a countable family $(F_n)_{n \in \mathbb N}$ of $\ell^2$-closed set whose interiors are empty and such that 
$$
\ell^1 \subseteq \bigcup_{n\in\mathbb N} F_n .
$$
What do you suggest? I tried with $B(0,n)=\{(x_k)_{k \in \mathbb N}: \sum_{k} \vert x_k\vert < n\}$ but I don't manage to prove - wheter it is true - that they are closed and with empty interior...

REPLY [5 votes]: $\def\norm#1{\left\|#1\right\|}$Let's take $\bar B_n = \{(x_k) \in \ell^1 \mid \norm x_1 \le n \}$. Let $y \in \ell^2\setminus \ell^1$, e.g. $y = (1/n)_n$, then for each $x \in \bar B_n$ and each $\epsilon > 0$, $x + \epsilon y \not\in \bar B_n \subseteq \ell^1$. So $\bar B_n$ has empty interior. It remains to prove the closedness. So let $x^k \in \bar B_n$ for $k \in \mathbb N$ and $x \in \ell^2$ with $\|x^k - x\|_2 \to 0$. 
Then, as $\ell^2$-convergence implies pointwise convergence
\begin{align*}
  \norm x_1 &= \sum_i |x_i|\\
    &= \lim_I \sum_{i=1}^I |x_i|\\
    &= \lim_I \sum_{i=1}^I\lim_k |x^k_i|\\
    &= \lim_I \lim_k \sum_{i=1}^I |x^k_i|\\
    &\le \limsup_I \limsup_k \norm{x^k}_1\\
    &\le n.
\end{align*}
So $x \in \bar B_n$ and we are done.