Is the union of all $l^p$ spaces meagre in $l^\infty$?

Question

Is the union of all $l^p$ spaces meagre in $l^\infty$? i.e. is $ \bigcup_{p=1}^\infty l^p$ meagre?

I am revisiting this variety of math after a long break so help is appreciated. Please correct anything I get wrong too.

I know that a space is meagre if it can be written as a countable union of nowhere dense sets, and that a set is nowhere dense if it is closed and has no interior points.

My attempts/ideas:

(1) I seem to remember that $l^1 \subset l^2 \subset l^3 \subset l^4 \subset ...\subset l^\infty$, but maybe I'm wrong about this? I can't seem to find it anywhere or prove it either. Is this true? If this were the case then it would be enough to show that $\bigcup_{p=1}^\infty l^p=l^\infty$ is meagre. Then I'm stuck again... how do I show that $l^\infty$ is meagre?

(2) There is already a countable union (of $l^p$s) here (do I have to show that it is countable? This is obvious isn't it?), so that makes me think that it would be good if I could show that each $l^p$ is nowhere dense. How could I do this?

Any critiques/ideas?

Answer 1

Observe that if $$x\in \bigcup_{p\in[1,\infty)}\ell^p,$$ then $x\in \ell^p$ for some $1\leq p <\infty$. thus $\sum |x_n|^p<\infty$. In particular $|x_n|\rightarrow 0$, so $$\bigcup_{p\in[1,\infty)}\ell^p\subset c_0.$$

Answer 2

Complementing Jonas answer, the sequence with terms $ x_k = \dfrac{1}{\ln(k+1)}$ for all $k\geq 1$ is in $c_0$, but it's not in any of the $l^{p}$ sets. Suppose, by contradiction, that $(x_k)_{k\in\mathbb{N}} \in l^{p}$. The idea to prove it is using Hölder's inequality and get the inequality, for each $k\in\mathbb{N}$, $$\sum^{k}_{i=1} \dfrac{1}{ (1+k) \ln(k+1)}\leq \left(\sum^{k}_{i=1} \dfrac{1}{ (1+k)^{q}}\right)^{1/q}\left( \sum^{k}_{i=1} \dfrac{1}{\ln(k+1)^p} \right)^{1/p}$$ with $\dfrac{1}{p}+\dfrac{1}{q}=1$, which would, erroneously, mean that the sequence of terms $ y_k = \dfrac{1}{(k+1)\ln(k+1)}$ belongs to $l^{1}.$

The argument to derive that $(y_k)_{k\in\mathbb{N}}$ is not in $l^{1}$ is to realize that, for each $k\in\mathbb{N}$, $$ \int^{k}_{1} \dfrac{1}{(\tau+1)\ln(\tau+1)} d\tau = \ln(\ln(k+1))- \ln(\ln(2)) \leq \sum^{k}_{i=1} \dfrac{1}{(k+1)\ln(k+1)}, $$ using the integral test for convergence.