Take the 2-minute tour ×  Prove that $\ell_1$ is first category
in $\ell_2$.  I tried to solve this, but had no idea about the
approach. Any suggestions are helpful. Thanks in advance.
share|improve this question  1 Answer 1  Write $F_n:=\{x\in
\ell^2,\sum_{j=1}^{+\infty}|x_j|\leq n\}$. Then
$\ell^1=\bigcup_{n\geq 1}F_n$. $F_n$ is closed in $\ell^2$, as
if $\{x^{(k)}\}$ is a sequence which lies in $F_n$ and converges
to $x$ in $\ell^2$; we have for an integer $N$ that $$\sum_{j=1}
^N|x_j|\leq\lim_{k\to\infty}\sum_{j=0}^N|x_j^{(k)}|\leq n,$$
which gives $x\in F_n$.  $F_n$ has an empty interior in
$\ell^2$. Otherwise, if $B_{\ell^2}(x,r)\subset F_n$, then for
each $y\in \ell²$, we would have $\frac{r}{2\lVert
y\rVert_2}y+x\in F_n$, hence $\frac{r}{2\lVert y\rVert_2}y\in
F_{2n}$. This gives that $\lVert y\rVert_1\leq C\lVert
y\rVert_2$ for an universal constant $C$, which is not possible.
share|improve this answer You are right, my bad. –  Martin
Argerami Oct 3 '12 at 18:34  Your Answer