I remember seeing this some time ago, but I can't find the examples anywhere.
Recall that if $p<q$, then $\ell^p\subseteq\ell^q$ and $L^q[0,1]\subseteq L^p[0,1]$. So we can ask ourselves if any of the equalities $$\ell^p=\bigcap_{q>p}\ell^q,\quad \ell^q=\bigcup_{p<q}\ell^p,\quad L^p[0,1]=\bigcup_{q>p}L^q[0,1],\quad L^q[0,1]=\bigcap_{p<q}L^p[0,1]$$ is true. If I remember it correctly, none of these is true, but I simply managed to find a counterexample for the first one: the sequence $(n^{-1/p})$ belongs to $\bigcap_{q>p}\ell^q$ but not to $\ell^p$.
Moreover, I recall that $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$ do not satisfy any inclusion relation.
What would be counter-examples to those equalities? Thank you.
For $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$, suppose $p < q$. Let $f(x) = x^{-2/(p+q)}$ for $x > 0$ and $0$ otherwise. Let $f_p(x) = f(x)$ if $x < 1$ and $0$ otherwise, and $f_q(x) = f(x)$ if $x > 1$ and $0$ otherwise. Then $f_p\in L^p \setminus L^q$ and $f_q \in L^q \setminus L^p$.
For $L^q[0,1]$ as a intersection of $L^p$ with $p < q$. The function $x^{-1/q}$ is an example.
For $L^q[0,1]$ as a union of $L^p$ with $p > q$, consider the function $(x (\log x)^2)^{-1/q}$.
For $\ell^q$ as a union of $\ell^p$ with $p < q$, I think the sequence $[n (\log (3+n))^2]^{-1/q}$ should work, but it needs to be checked. If not, something very similar would work.