$l^p$ is of first category in $l^q$ if $1 \leq p < q< \infty$ Let $1\leq p<q<+\infty$. Let $B_n=\{<x_k>\in l^q: \sum_k |x_k|^p\leq n\}$. Want to show: $B_n$ is closed and nowhere dense in $l^q$. Thus $l^p$ is of first category as a subset of $l^q$.
To show $B_n$ doesn't have interior points seems easy, since you can always construct a sequence of numbers that has arbitrarily small $l^q$-norm but diverges in $l^p$. But how to show it's closed? How to show that the $l^q$-llimit of a sequence of elements in $l^p$ is also in $l^p$? I have no idea how to estimate the $l^p$-norm of the limit..
 A: Partly for notational convenience and also partly to clarify that this isn't something special about $l^p$ and $l^q$, let us identify $l^p$ with $L^p(\mathbb{N},c)$ and $l^q$ with $L^q(\mathbb{N},c)$, where $c$ is the counting measure. For general $(X,\mu)$ we can prove the same statement for $L^p(X,\mu)$ and $L^q(X,\mu)$ as follows.
Let $B_n = \{ x \in L^q : \| x \|_p \leq n \}$. Suppose $x_m$ is a sequence in $B_n$ which converges in the sense of $L^q$ to $x$. To show $B_n$ is closed, it is enough to show that $x \in B_n$. To show $x \in B_n$, we need to show $x \in L^q$ and $\| x \|_p \leq n$. The first part follows because $x$ is the $L^q$ limit of a sequence. 
To prove the second part, apply the $L^q$ convergence to extract a subsequence $x_{m_k}$ which converges almost everywhere. Now use Fatou's lemma to get that $\| x \|_p \leq \liminf \| x_{m_k} \|_p \leq n$. 
So $x \in B_n$, and so $B_n$ is closed in the sense of $L^q$.
A: We first prove that $B_{n}$ is closed.Let $\{x_{n}\}_{n \in N}$ be a sequence in $B_{n}$,converges with the limit $x \in l_{q}$ and considering $n=1$ without loss of generality.
For a fixed $n \in N$ and again then for each $i \in N$;we have,
$\mid{x^{(i)}-x_{n}^{(i)}}\mid \leq \left\|x-x_{n}\right\|_{q} \to 0$ as $n\to \infty$,so that $x_{n}^{i} \to x^{i}$
 as $n \to \infty$.But then,
$\sum_{i=1}^{J}\mid{x^{(i)}}\mid^{p}=lim_{n\to \infty}\sum_{i=1}^{J}\mid{x_{n}^{(i)}}\mid^{p} \leq lim_{n \to \infty}sup \left\|x_{n}\right\|_{p}^{p} \leq 1$.
We conclude that $x\in l^{p}$ with $\left\|x\right\|_{p}\leq 1$ and thus $x\in B_{n}$.So $B_{n}$ is closed.
Secondly,we prove that $B_{n}$ has an empty interior in $l_{q}$.Pick any $x \in B_{n}$.We will show that $x$ is not an interior point of $B_{n}$.
We can pick $y\in l^{q} \setminus l^{p}$.Let $r$>$ 0$ and set ${\overline x}:= \frac{r}{2\left\|y\right\|_{q}}y+x$.
Note that $\overline x \in l_{q}$ as a sum of two elements of $l_{q}$,but $\overline x \not\in l_{p}$.Indeed,if $\overline x \in l_{p}$,then also $y=\frac{2 \left\|y \right\|_{q}(\overline x - x)}{r} \in l_{p}$ Contradicting,our fact,$y \not\in l_{p}$.We conclude that $\overline x \in l_{q} \setminus l_{p}$,where $q>p$.
Note that,
$\left\| \overline x - x\right\|_{q}=\frac{r}{2} $<$r$.
However,since $\overline x \not\in l_{p}$,we do not have $x\in B_{n}$.Thus the ball $B(x,r)$ is not contained in $B_{n}$.As $r>0$ was arbitrary,we conclude that x is not an interior point of $B_{n}$ and thus $B_{n}$ has an empty interior.Thus we have shown that $B_{n}$ is a nowhere dense set in $l_{q}$.$l_{p}$ can be expressed as a countable union of nowhere dens sets.Thus $l_{p}$ is of first category in $l_{q}$
