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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...

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marked as duplicate by Davide Giraudo, tomasz, sdcvvc, Martin Sleziak, Asaf Karagila Oct 13 '12 at 12:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

math.stackexchange.com/questions/206641/… will help you. –  Davide Giraudo Oct 12 '12 at 14:22

1 Answer 1

up vote 5 down vote accepted

$\def\norm#1{\left\|#1\right\|}$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.

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